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CALCULUS 


MODERN  MATHEMATICAL  TEXTS 

EDITED  BY 

Charles  S.  Slighter 


ELEMENTARY  MATHEMATICAL  ANALYSIS 
By  Charles  S.  Slighter 
490  pages,  5  x  IM,  Illustrated. 

MATHEMATICS  FOR  AGRICULTURAL 
STUDENTS 
By  Henry  C.  Wolff 
311  pages,  5  x  7J.i,  Illustrated . 

CALCULUS 
By  Herman  W.  March  and  Henry  C.  "Wolff 
360  pages,  5  x  TVi,  Illustrated . 


MODERN    MATHEMATICAL    TEXTS 
Edited  by   Charles  S.   Slighter 


CALCULUS 


BY 
HERMAN  W.  MARCH,  Ph.  D. 


ASSOCIATE   PKOFE880R  OP   MATHEMATICS 
UNIVEKSITY   OF  WISCONSIN 


AND 


HENRY  C.  WOLFF,  Ph.  D. 

PBOFESSOR  OF  MATHEMATICS 
DREXEIi  INSTITUTE 


First  Edition 
Sixth  Impression 


McGRAW-HILL  BOOK  COMPANY,  Inc. 
NEW  YORK:  370  SEVENTH  AVENUE 

LONDON:  6  &  8  BOUVERIE  ST.,  E.  C.  4 
1917 


Copyright,  1917,  by  the 
McGraw-Hill  Book  Company,  Inc. 


THE  MAPLE  PRESS  YORK  PA 


Geolo^^ 
Library 


PREFACE 


One  of  the  purposes  of  the  elementary  working  courses  in  mathe- 
matics of  the  freshman  and  sophomore  years  is  to  exhibit  the  bond 
that  unites  the  experimental  sciences.  "The  bond  of  union 
among  the  physical  sciences  is  the  mathematical  spirit  and  the 
mathematical  method  which  pervade  them."  For  this  reason,  the 
applications  of  mathematics,  not  to  artificial  problems,  but  to  the 
more  elementary  of  the  classical  problems  of  natural  science,  find 
a  place  in  every  working  course  in  mathematics.  This  presents 
probably  the  most  difficult  task  of  the  text-book  writer, — namely, 
to  make  clear  to  the  student  that  mathematics  has  to  do  with  the 
laws  of  actual  phenomena,  without  at  the  same  time  undertaking 
to  teach  technology,  or  attempting  to  build  upon  ideas  which  the 
student  does  not  possess.  It  is  easy  enough  to  give  examples  of 
the  application  of  the  processes  of  mathematics  to  scientific  prob- 
lems; it  is  more  difficult  to  exhibit  by  these  problems,  how,  in 
mathematics,  the  very  language  and  methods  of  thought  fit 
naturally  into  the  expression  and  derivation  of  scientific  laws  and  of 
natural  concepts. 

It  is  in  this  spirit  that  the  authors  have  endeavored  to  develop 
the  fundamental  processes  of  the  calculus  which  play  so  important 
a  part  in  the  physical  sciences;  namely,  to  place  the  emphasis  upon 
the  mode  of  thought  in  the  hope  that,  even  though  the  student  may 
forget  the  details  of  the  subject,  he  will  continue  to  apply  these 
fundamental  modes  of  thinking  in  his  later  scientific  or  technical 
career.  It  is  with  this  purpose  in  mind  that  problems  in  geometry, 
physics,  and  mechanics  have  been  freely  used.  The  problems 
chosen  will  be  readily  comprehended  by  students  ordinarily  taking 
the  first  course  in  the  calculus. 

A  second  purpose  in  an  elementary  working  course  in  mathe- 
matics is  to  secure  facility  in  using  the  rules  of  operation  which 
must  be  applied  in  calculations.  Of  necessity  large  numbers  of 
drill  problems  have  been  inserted  to  furnish  practice  in  using  the 


104^284 


vi  PREFACE 

rules.  It  is  hoped  that  the  solution  of  these  problems  will  be  re- 
garded by  teacher  and  student  as  a  necessary  part  but  not  the 
vital  part  of  the  course. 

While  the  needs  of  technical  students  have  been  particularly  in 
the  minds  of  the  authors,  it  is  believed  that  the  book  is  equally 
adapted  to  the  needs  of  any  other  student  pursuing  a  first  course 
in  calculus.  The  authors  do  not  believe  that  the  purposes  of 
courses  in  elementary  mathematics  for  technical  students  and  for 
students  of  pure  science  differ  materially.  Either  of  these  classes 
of  students  gains  in  mathematical  power  from  the  type  of  study 
that  is  often  assumed  to  be  fitted  for  the  other  class. 

In  agreement  with  many  others,  the  book  is  not  divided  into  two 
parts.  Differential  Calculus  and  Integral  Calculus.  Integration 
with  the  determination  of  the  constant  of  integration,  and  the 
definite  integral  as  the  limit  of  a  sum,  are  given  immediately  fol- 
lowing the  differentiation  of  algebraic  functions  and  before  the 
differentiation  of  the  transcendental  functions.  With  this  arrange- 
ment many  of  the  most  important  applications  of  the  calculus 
occur  early  in  the  course  and  constantly  recur.  Further,  with  this 
arrangement,  the  student  is  enabled  to  pursue  more  advantageously 
courses  in  physics  and  mechanics  simultaneously  with  the  calculus. 

The  attempt  has  been  made  to  give  infinitesimals  their  proper 
importance.  In  this  connection  Duhamel's  Theorem  is  used  as  a 
valuable  working  principle,  though  the  refinements  of  statement 
upon  which  a  rigorous  proof  can  be  based  have  not  been  given. 

The  subjects  of  center  of  gravity  and  moments  of  inertia  have 
been  treated  somewhat  more  fully  than  is  usual.  They  are  par- 
ticularly valuable  in  emphasizing  the  concept  of  the  definite 
integral  as  the  limit  of  a  sum  and  as  a  mode  of  calculating  the 
mean  value  of  a  function.  Sufficient  solid  analytic  geometry  is 
given  to  enable  students  without  previous  knowledge  of  this  sub- 
ject to  work  the  problems  involving  solids.  In  the  last  chapter 
simple  types  of  differential  equations  are  taken  up. 

The  book  is  designed  for  a  course  of  four  hours  a  week  through- 
out the  college  year.  But  it  is  easy  to  adapt  it  to  a  three-hour 
course  by  suitable  omissions. 

The  authors  are  indebted  to  numerous  current  text-books  for 
many  of  the  exercises.     To  prevent  distracting  the  student's  at- 


PREFACE  vii 

tention  from  the  principles  involved,  exercises  requiring  compli- 
cated reductions  have  been  avoided  as  far  as  possible. 

The  book  in  a  preliminary  form  has  been  used  for  two  years 
with  students  in  the  College  of  Engineering  of  the  University  of 
Wisconsin.  Many  improvements  have  been  suggested  by  our 
colleagues,  Professor  H.  T.  Burgess,  Messrs.  E.  Taylor,  T.  C.  Fry, 
J.  A.  Nyberg,  and  R.  Keffer.  Particular  acknowledgment  is  due 
to  the  editor  of  this  series.  Professor  C.  S.  Slichter,  for  suggestions 
as  to  the  plan  of  the  book  and  for  suggestive  criticism  of  the  manu- 
script at  all  stages  of  its  preparation. 

The  authors  will  feel  repaid  if  a  little  has  been  accomplished 
toward  presenting  the  calculus  in  such  a  way  that  it  will  appeal  to 
the  average  student  rather  as  a  means  of  studying  scientific  prob- 
lems than  as  a  collection  of  proofs  and  formulas. 

Unfversity  op  Wisconsin,  Herman  W.  March, 

November  6,  1916.  Henry  C,  Wolff. 


CONTENTS 

Page 

Preface     v 

Introduction 1 

1.  Constant.     Variable.     Function  .    .  - 1 

2.  The  Power  Function 1 

3.  The  Law  of  the  Power  Function       3 

4.  Polynomials.     Algebraic  Function 4 

5.  Transcendental  Functions      " . 6 

6.  Translation 6 

7.  Elongation  or  Contraction,  or  Orthographic  Projection,  of  a 
Curve .  7 

8.  Shear 8 

9.  The  Function  a' 9 

10.  The  Function  sin  x      9 

11.  The  Functions  p  =  a  cos  0,  p  =  b  sin  0,  and  p  =  a  cos  d   + 

6  sin  e 9 

12.  Fundamental  Transformations  of  Functions 10 

CHAPTER  I 

Derivative 

13.  Increments 13 

14.  The  Function  y  =  x^ 16 

15.  Slope  of  the  Tangent 17 

16.  Maxima  and  Minima 20 

17.  Derivative 21 

18.  Velocity  of  a  Falling  Body 22 

19.  Illustrations      23 

20.  Illustrations      23 

CHAPTER  II 

Limits 

21.  Definition 27 

22.  Notation 29 


X  CONTENTS 

Section  Page 

23.  Infinitesimal 29 

24.  Theorems  on  Limits 30 

25.  The  Indeterminate  Form  ^ 30 

26.  Continuous  and  Discontinuous  Functions 31 

CHAPTER  III 

Thf  Power  Function 

27.  The  Power  Function 33 

28.  The  Derivative  of  x»       33 

29.  The  Derivative  of  ax" 35 

30.  Rate  of  Change  of  ax" 37 

31.  The  Derivative  of  the  Sum  of  a  Function  and  a  Constant  .  39 

32.  The  Derivative  of  au" 41 

33.  The  Derivative  of  u",  n  a  Positive  Fraction 42 

34.  The  Derivative  of  a  Constant 43 

35.  The  Derivative  of  the  Sum  of  Two  Functions 43 

36.  Differentiation  of  ImpHcit  Functions       44 

37.  Anti-derivative.     Integration 46 

38.  Acceleration      52 

CHAPTER  IV 

DiPFEKENTIATION  OF  AlGBBRATION  FUNCTIONS 

39.  The  Derivative  of  the  Product  of  a  Constant  and  a  Variable.  56 

40.  The  Derivative  of  the  Product  of  Two  Functions 57 

41.  The  Derivative  of  the  Quotient  of  Two  Functions    ....  58 

42.  The  Derivative  of  w",  n  negative 59 

43.  Maximum  and  Minimum  Values  of  a  Function 00 

44.  Derivative  of  a  Function  of  a  Function 66 

45.  Inverse  Functions 67 

46.  Parametric  Equations 67 

47.  Lengths  of  Tangent,  Normal,  Subtangent,  and  Subnormal  68 

CHAPTER  V 
Second  Derivative.     Point  of  Inflection 

48.  Second  Derivative.     Concavity 70 

49.  Points  of  Inflection 71 


CONTENTS  xi 

CHAPTER  VI 

Applications 
Section  Page 

50.  Area  Under  Curve:  Rectangular  Coordinates 75 

51.  Work  Done  by  a  Variable  Force 77 

52.  Parabolic  Cable 82 

53.  Acceleration 83 

54.  The  Path  of  a  Projectile 85 

CHAPTER  VII 

Infinitesimals.     Differentials.     Definite  Integrals 

55.  Infinitesimals 88 

56.  j;'!'.'^ •    ■    ■    ■         88 

lim    tan  a    lim   tan  a 
^'-  a  =  Q~^'(x^Q^vr~a. ^^ 

58.  ^^    l^^^ 89 

a:  =  0  a 

59.  Order  of  Infinitesimals 90 

60.  Theorem 93 

61.  Differentials 95 

62.  Formulas  for  the  Differentials  of  Functions 96 

63.  Differential  of  Length  of  Arc :  Rectangular  Coordinates  .    .  100 

64.  The  Limit  of  2;/(a;)Aa;      102 

65.  Definite  Integral      103 

66.  Duhamel's  Theorem 105 

•67.  Work  Done  by  a  Variable  Force 107 

68.  Volume  of  a  Solid  of  Revolution 109 

69.  Length  of  Arc:  Rectangular  Coordinates     . Ill 

70.  Area  of  a  Surface  of  Revolution 112 

71.  Element  of  Integration       113 

72.  Water  Pressure 114 

73.  Airthmetic  Mean 116 

74.  Mean  Value  of  a  Function 117 

CHAPTER  VIII 

Circular  Functions.     Inverse  Circular  Functions 

75.  The  Derivative  of  sin  m      122 

76.  The  Derivative  of  cos  u,  tan  m,  cot  u,  sec  m,  and  esc  u    .    .124 


xii  CONTENTS 

Section  Page 

77.  The  Derivative  of  the  Inverse  Circular  Functions    ....   131 

78.  Velocity  and  Acceleration      136 

79.  Angular  Velocity  and  Acceleration       139 

80.  Simple  Harmonic  Motion 140 

81.  The  Simple  Pendulum 141 

CHAPTER  IX 

Exponential  and  Logarithmic  Functions 

82.  The  Derivative  of  the  Exponential  and  Logarithmic  Func- 
tions         145 

83.  Logarithmic  Differentiation 153 

84.  Compound  Interest  Law 155 

85.  Relative  Rate  of  Increase 159 

86.  Hyperbolic  Function 159 

87.  Inverse  Hyperbolic  Functions 160 

88.  The  Catenary       161 

CHAPTER  X 

Maxima  and  Minima 

89.  The  Maximum  or  Minimum  of  y  =  ax^  +  fix  +  y  .    .    .    .  165 

90.  The  Function  a  cos  x  +  b  sin  x 165 

91.  The  Function  mx  ±  \^a^  -  s" 168 

92.  Ma.xima  and  Minima  by  Limits  of  Curve 169 

93.  Maxima  and  Minima  Determined  by  the  Derivative   .    .    .  169 

94.  Second-derivative  Test  for  Maxima  and  Minima 172 

95.  Study  of  a  Function  by  Means  of  its  Derivatives 172 

96.  Applications  of  Maxima  and  Minima      174 

CHAPTER  XI 

Polar  Coordinates 

97.  Direction  of  Curve  in  Polar  Coordinates 178 

98.  Differential  of  Arc :  Polar  Coordinates 181 

99.  Area:  Polar  Coordinates 183 


CONTENTS  xiii 

CHAPTER  XII 

Integration 

Section  Page 

100.  Formulas       185 

101.  Integration  of  Expressions  Containing  ax^  +  bx  -\-  c,  by 
Completing  the  Square       190 

102.  Integrals  Containing  Fractional  Powers  of  a:  or  of  a  +  fca;  .  192 

103.  Integrals  of  Powers  of  Trigonometric  Functions 193 

104.  Integration  of  Expressions  Containing  \/a^  —  x^,'\/a'^  +  x^, 
'\/x^  —  o*,  by  Trigonometric  Substitution 196 

105.  Change  of  Limits  of  Integration 199 

106.  Integration  by  Parts 201 

.07.Tho.„tegraU/e«si„n...,/.«c«„... .202 

108.  J  sm3  xdx 204 

109.  Wallis'  Formulas 204 

,^-,    .,   ,         ,.         ,    Ta  sin  X  +  6  cos  X  , 

110.  Integration  of    I  — -. -. — j ax 209 

J  c  sm  x  -\-  a  cos  x 

111.  Partial  Fractions 211 

CHAPTER  XIII 

Applications  of  the  Process  of  Integration.     Improper 
Integrals 

112.  Summary  of  Applications 208 

113.  Improper  Integrals      223 

114.  Improper  Integral:  Infinite  Limits       226 

CHAPTER  XIV 
Solid  Geometry 

115.  Coordinate  Axes.     Coordinate  Planes 228 

116.  The  Distance  Between  Two  Points      230 

117.  Direction  Cosines  of  a  Line 230 

118.  Angle  Between  Two  Lines 231 

119.  The  Normal  Form  of  the  Equation  of  a  Plane 232 

120.  The  Equation  Ax  +  By  +  Cz  ==  D 233 

121.  Intercept  Form  of  the  Equation  of  a  Plane    . 234 

122.  The  Angle  Between  Two  Planes 236 

123.  Parallel  and  Perpendicular  Planes 236 


xiv  CONTENTS 

Section  Page 

124.  The  Distance  of  a  Point  From  a  Plane 237 

125.  Symmetrical  Form  of  the  Equation  of  a  Line 238 

126.  Surface  of  Revolution 240 

127.  Quadric  Surfaces      242 

128.  Cylindrical  Surfaces 244 

129.  Partial  Derivatives 246 

130.  Partial  Derivatives  of  Higher  Order 247 

CHAPTER  XV 

SuccEssrvE  Integration.    Center  of  Gravity.     Moment 
OF  Inertia 

131.  Introduction 250 

132.  Illustration  of  Double  Integration 251 

133.  Area  by  Double  Integration :  Rectangular  Coordinates .    .    .  253 

134.  Geometrical  Meaning  of  the  Definite  Double  Integral    .    .    .  255 

135.  Area:  Polar  Coordinates 258 

136.  Volume  of  a  Solid:  Triple  Integration 260 

137.  Center  of  Mass.     Centroid 263 

138.  Centroid  Independent  of  the  Position  of  the  Coordinate 
Planes 264 

139.  Center  of  Gravity    . 264 

140.  Centroid  of  a  Continuous  Mass 265 

141.  Theorems  of  Pappus 274 

142.  Centroid:  Polar  Coordinates 276 

143.  Moment  of  Inertia  .    .    .   ' 277 

144.  Transfer  of  Axes      280 

145.  Moment  of  Inertia  of  an  Area 281 

146.  Moment  of  Inertia:  Polar  Coordinates 283 

147.  Moment  of  Inertia  of  a  Solid 284 

CHAPTER  XVI 

Curvature.     Evolutes.     Envelopes 

148.  Curvature 289 

149.  Curvature  of  a  Circle      290 

150.  Circle   of   Curvature.     Radius   of    Curvature.     Center   of 
Curvature 290 

151.  Formula  for  Curvature  and  Radius  of  Curvature:  Rectan- 
gular Coordinates 290 


CONTENTS  XV 

Section  Page 

152.  Curvature:  Parametric  Equations 292 

153.  Approximate  Formula  for  the  Curvature 293 

154.  Center  of  Curvature.     Evolute 294 

155.  Envelopes 296 

156.  The  Evolute  as  the  Envelope  of  the  Normals 300 

157.  Involutes       302 

CHAPTER  XVII 

Series.     Taylor's   and    Maclaurin's  Theorems.     Indeter- 
minate Forms 

158.  Infinite  Series 303 

159.  Rolle's  Theorem 305 

160.  Law  of  the  Mean 306 

161.  The  Extended  Law  of  the  Mean 307 

162.  Taylor's  Theorem  with  the  Remainder 308 

163.  Taylor's  and  Maclaurin's  Series 311 

164.  Second  Proof  for  Taylor's  and  Maclaurin's  Series   ....  313 

165.  Tests  for  the  Convergence  of  Series 315 

166.  Computation  of  Logarithms 320 

167.  Computation  of  tt 321 

168.  Relation  Between  the  Exponential  and  Circular  Functions  321 

169.  DeMoivre's  Theorem      323 

170.  Indeterminate  Forms      324 

CHAPTER  XVIII 
Total- Derivative.     Exact  Differential 

171.  The  Total  Derivative 328 

172.  Exact  Differential 332 

173.  Exact  Differential  Equations 335 

174.  Envelopes .336 

CHAPTER  XIX 

Differential  Equations 

175.  Differential  Equations        338 

176.  General  Solution.     Particular  Solution 338 

177.  Exact  Differential  Equation 339 

178.  Differential  Equations:  Variable  Separable 339 


xvi  CONTENTS 

Section  Page 

179.  Homogeneous  Differential  Equations 340 

180.  Linear  Differential  Equations  of  the  First  Order 342 

181.  Extended  Form  of  the  Linear  Differential  Equation     .    .    .   344 

182.  Applications      345 

183.  Linear  Differential  Equations  of  Higher  Order  with  Con- 
stant Coefficients  and  Second  Member  Zero 347 

184.  Au.xiliary  Equation  with  Equal  Roots 349 

185.  Auxiliary  Equation  with  Complex  Roots 350 

186.  Damped  Harmonic  Motion 351 

Index 154 


CALCULUS 


INTRODUCTION 

1.  Constant.  Variable.  Function.  1.  A  symbol  of  number 
or  quantity,  as  a,  to  which  a  fixed  value  is  assigned  throughout 
the  same  problem  or  discussion  is  called  a  constant. 

2.  A  symbol  of  number  or  quantity,  as  x,  to  which  a  succession 
of  values  is  assigned  in  the  same  problem  or  discussion  is  called  a 
variable. 

Example.  The  mass  or  weight  of  mercury  in  a  thermometer  is 
constant.  The  number  that  results  from  measuring  this  quantity 
(weight)  is  a  constant. 

The  volume  of  the  mercury  in  the  thermometer  is  variable. 
The  number  that  results  from  measuring  this  quantity  (volume) 
is  a  variable. 

3.  The  variable  y  is  said  to  be  a  function  of  the  variable  x  if, 
when  X  is  given,  one  or  more  values  of  y  are  determined. 

4.  X,  the  variable  to  which  values  are  assigned  at  will  is  called 
the  independent  variable,  or  the  argument  of  the  function. 

5.  y,  whose  values  are  thereby  determined,  is  called  the  de- 
pendent variable. 

6.  y  is  said  to  be  a  function  of  several  variables  u,  v,  w,  •  •  • 
if,  when  u,  v,  w,  •  •  •  are  given,  one  or  more  values  of  y  are 
determined. 

7.  The  variables  u,  v,  w,  '  •  •  ,  to  which  values  are  assigned 
at  will  are  called  the  independent  variables,  or  the  arguments  of  the 
function. 

Functions  of  a  single  variable  or  argument  are  represented 
by  symbols  such  as  the  following:  f{x),  F{x),  4>{x),  yp{x).  Func- 
tions of  several  arguments  are  represented  by  symbols  such  as 
f{u,  v,  w),  F{u,  v,  w),  0(w,  V,  w). 

2.  The  Power  Function.  $.  The  function  x",  where  n  is  a 
constant,  is  called  the  power  function. 

1 


CALCULUS 


(§2 


If  n  is  positive  the  function  is  said  to  be  of  the  parabolic  type, 
and  the  curve  representing  such  a  function  is  also  said  to  be  of  the 
parabolic  type.     If  w  =  2,  the  curve,  y  =  x"^,  is  a  parabola. 


-^y 


4-Li'' 


Fig.  1.— Curves  for  y  =  x",  n  =  1,  2,  3,  and  4. 

If  n  is  negative  the  function  x"  is  said  to  be  of  the  hyperbolic 
type,  and  the  curve  representing  such  a  function  is  also  said  to  be 


§3] 


INTRODUCTION 


of  the  hyperbolic  type.     If  n  =  —1,  the  curve,  y  =  x-^,  is  an 
equilateral  hyperbola. 

In  Figs.  1,  2,  3,  and  4,  curves  representing  ?/  =  x"  for  different 
values  of  n  are  drawn.  In  Fig.  1,  n  has  positive  integral  values; 
in  Fig.  2,  positive  fractional  values;  in  Fig.  3,  negative  integral 
values;  and  in  Fig.  4,  negative  fractional  values.     The  curves  for 


StY" 


Fig.  2. — Curves  for  y  =  x",  n  =  5,  5,  f,  and  |. 


y  =  x"  all  pass  through  the  point  (1,  1).  They  also  pass  through 
the  point  (0,  0)  if  n  is  positive.  If  n  is  negative,  they  do  not 
pass  through  (0,  0).  In  the  latter  case  the  coordinate  axes  are 
asymptotes  to  the  curves. 

3.  The  Law  of  the  Power  Function.  9.  In  any  power  function, 
if  X  changes  by  a  fixed  multiple,  y  also  changes  by  a  fixed  multiple. 

The  same  law  can  be  stated  as  follows: 


CALCULUS 


m 


10.  In  any  power  function,  if  x  increases  by  a  fixed  percent, 
y  also  increases  by  a  fixed  percent. 

The  preceding  statements  are  also  equivalent  to  the  fol- 
lowing: 

11.  In  any  power  function,  if  x  runs  over  the  terms  of  a 
geometrical  progression,  then  y  also  runs  over  the  terms  of  a 
geometrical  progression. 


Fig.  3. — Curves  for  y  =  x",  n  =  —1,  —2,  and  —3. 

4.  Polynomials.  Algebraic  Function.  12.  A  polynomial  in 
a;  is  a  sum  of  a  finite  number  of  terms  of  the  form  ax",  where  a  is  a 
constant  and  n  is  a  positive  integer  or  zero.    For  example: 

ax^  +  bx^  -^  ex  -\-  d. 

13.  A  polynomial  in  x  and  y  is  a  sum  of  a  finite  number  of 


INTRODUCTION 


terms  of  the  form  ox'»i/'»,  where  a  is  a  constant  and  m  and  n  are 
positive  integers  or  zero.     For  example: 

axhj^  +  ^xy^  +  cx^  -{•  dy  -{•  e. 
1  i.  Functions  of  a  variable  x  which  are  expressed  by  means  of  a 
finite  number  of   terms  involving   only   constant  integral   and 

Y 


w=--|- 


Fig.  4. — Curves  for  y  =  x^,  n  ■=  —  |,  —  5,  — f,  and  — |. 

fractional  powers  of  x  and  of  polynomials  in  x  are  included  in  the 
class  of  functions  known  as  algebraic  Junctions^  of  x.     For  example : 

(a)  x\ 


(d)  ^,  +  1  +  1- 


(b)  a;3  +  (2a; 


3)i. 


a;''   ■  X 
ie)  X  +  5  + 


(c)   Vx^  +  4x  +  7  +  4x  +  5.       (/) 


■\/x  —  7 
3a;2  +  5a;  +  7 


x3  _  3a;  +  2 

1  A  function  of  x  defined  by  the  equation  F(,x,  y)  =  0,  where  F{x,  y)  is  a  polynomial 
in  X  and  y,  is  an  algebraic  function  of  x.  For  example,  y  =  \/xs  +  2  is  an  algebraio 
function  of  x.     For  by  squaring  and  transposing  we  obtain 

2/2  -  x2  -  2  =  0, 
in  which  the  first  member  is  a  polynomial  in  x  and  y. 


6  CALCULUS  f§5 

15.  An  algebraic  function  is  said  to  be  rational  if  it  can  be 
expressed  by  means  of  only  integral  powers  of  x  together  with 
constants. 

Rational  algebraic  functions  are  divided  into  two  classes: 
rational  integral  functions  and  rational  fractional  functions. 

16.  A  rational  integral  function  of  x  is  a  polynomial  in  x. 

17.  A  rational  fractional  function  is  a  quotient  of  two  poly- 
nomials in  X. 

It  is  usually  desirable  to  reduce  rational  fractional  functions 
of  a:  to  a  form  in  which  the  numerator  is  of  lower  degree  than 
the  denominator.  This  can  always  be  done  by  performing  long 
division. 

X  +  3  2 

Thus    y  =       .   1  is  equivalent  to  y  =  1  -\ ^rjy,    and 

Sx^  +  5x  +  7  .         .     ,     ,,  o   ,       Ux  +  1 

5.  Transcendental  Functions.  The  circular  (or  trigonometric), 
the  logarithmic,  and  the  exponential  functions  are  included  in  the 
class  of  functions  known  as  transcendental^  functions. 

6.  Translation.     If,  in  the  equation  of  a  curve 

fix,  y)  =  0, 
X  is  replaced  by  (x  —  a),  the  resulting  equation, 

fix  -  a,  y)  =  0, 

represents  the  first  curve  translated  parallel  to  the  axis  of  a;  a 
distance  a;  to  the  right  if  a  is  positive;  to  the  left  if  a  is 
negative. 

If  y  is  replaced  by  (y  —  /3)  the  resulting  equation, 

fix,  2/  -  /3)  =  0, 

represents  the  original  curve  translated  parallel  to  the  axis  of  y 
a  distance  j8;  up  if  /3  is  positive;  down  if  /3  is  negative.  Thus 
V  =  (x  +  3)^  —  4  is  the  parabola  y  =  x-  translated  three  units 
to  the  left  and  four  units  down.     See  Fig.  5. 

*  All  functions  which  are  not  algebraic  functions  as  defined  by  the  footnote  on  p. 
6  are  transcendental  functions. 


§7] 


INTRODUCTION 


7.  Elongation  or  Contraction,  or  Orthographic  Projection,  of  a 

X 


The  substitution  of  -  for  x  in  the  equation  of  any  locus 


Locus 

multiplies  all  of  the  abscissas  by  a. 


X-f 


This  transformation  can  be  considered  as  the  orthographic 
projection  of  a  curve  lying  in  one  plane  upon  another  plane,  the 
two  planes  intersecting  in  the  axis  oiy.  If  a  <  1  the  second  curve 
is  the  projection  of  the  former  curve  upon  a  second  plane  through 


the  F-axis  and  making  an  angle  a,  whose  cosine  is  equal  to  a, 
with  the  first  plane.  If  a  >  1,  the  first  curve  is  the  projection  of 
the  second  when  the  cosine  of  the  angle  between  their  planes 

.     1 

IS  -. 


CALCULUS 


y 


Similarly  the  substitution  of  -  for  y  in  the  equation  of  a  locus 

multiplies  the  ordinates  by  a.  The  interpretation  from  the 
standpoint  of  orthographic  projection  is  evident  from  what  has 
just  been  said.     See  Figs.  6  and  7. 

8.  Shear.  The  curve  y  =  f{x)  +  mx  is  the  curve  y  =  f{x) 
sheared  in  the  line  y  =  mx  in  such  a  way  that  the  y-intercepts 
remain  unchanged.  Every  point  on  the  curve  y  =  f{x)  to  the 
right  of  the  F-axis  is  moved  up  (down  if  m  is  negative)  a  distance 
proportional  to  its  abscissa;  and  every  point  to  the  left  of  the  Y- 


axis  is  moved  down  (up  if  m  is  negative)  a  distance  proportional 
to  its  abscissa.    The  factor  of  proportionality  is  m. 

In  general  a  curve  is  changed  in  shape  by  shearing  it  in  a  line. 
The  parabola  is  an  exception  to  this  rule. 

Thus  y  =  dx^  sheared  in  the  line  y  =  mx  becomes 

y  =  ax^  +  mx, 
or 

y  =  ^[''  +  2a)   -ia' 
This  may  also  be  considered  as  the  result  of  translating  the 


§11] 


INTRODUCTION 


9 


4M  tfl 

original  curve  by  the  amounts  —  ^  and  —  j-  in  the  x  and  ij  direc- 
tions, respectively.  Hence,  by  shearing,  the  parabola  y  =  az^ 
is  merely  translated. 

9.  The  Function  a*.  In  Fig.  8  are  given  the  graphs  of  ?/  =  a*, 
for  the  values  a  =  1,  2,  and  3.  By  reflecting  these  curves  in  the 
line  y  =  X  we  have  the  corresponding  curves  for  y  =  logo  x. 

The  exponential  function  y  =  a"  has  the  property  that  if  x  is 
given  a  series  of  values  in  arithmetical  progression  the  corre- 
sponding values  of  y  are  in  geometrical  progression. 


Fig.  8. — Curves  for  y  =  a*,  a  =  1,  2,  and  3. 


10.  The  Function  sin  x.  The  function  y  =  sinx  is  repre- 
sented in  Fig.  53. 

11.  The  Functions  p  =  a  cos  9,  p  =  6  sin  ^,  and  p  =  a  cos  6  + 
b  sin  6.  The  function  p  =  a  cos  d  is  the  circle  OA,  Fig.  9,  and 
p  =  bsin  d  is  the  circle  OB,  Fig.  9.  The  function  p  =  a  cos  6  + 
6  sin  d  can  be  put  in  the  form  p  =  R  cos  {8  —  a),  where 


10 


CALCULUS 


[§12 


R  =  \/a^  +  6S  and  where  cos  «  =  n  and  sin  a  =  „  •    This  function 

is  represented  by  a  circle,  Fig.  9,  passing  through  the  pole,  with 
diameter  equal  to  R,  and  with  the  angle  AOC  equal  to  a.  The 
maximum  value  of  the  function  is  R  and  the  minimum  value  is 
-R. 

12.  Fimdamental  Transformations  of  Functions.  It  is  valuable 
to  formulate  the  transformations  of  simple  functions,  that  most 
commonly  occur,  in  terms  of  the  effect  that  these  transformations 
have  upon  the  graphs  of  the  functions.  The  following  list 
of  theorems  on  loci  contains  useful  facts  concerning  these 
transformations : 


Fig.  9. 


THEOREMS  ON  LOCI 

I.  If  X  be  replaced  by  {—x)  in  any  equation  containing  x  and 
y,  the  new  graph  is  the  reflection  of  the  former  graph  in  the  F-axis. 

II.  If  y  be  replaced  by  {—y)  in  any  equation  containing  x  and 
y,  the  new  graph  is  the  reflection  of  the  former  graph  in  the  X-axis. 

III.  If  X  and  y  be  interchanged  in  any  equation  containing 
x  and  y,  the  new  graph  is  the  reflection  of  the  former  graph  in  the 
line  y  =  X. 

IV.  Substituting  /  - )  for  x  in  the  equation  of  any  locus  multipUes 
all  abscissas  l)y  a. 


§12]  INTRODUCTION  11 

V.  Substituting  L\  for  y  in  the  equation  of  any  locus  multiplies 

all  ordinates  of  the  curve  by  b. 

VI.  If  (x  —  a)  be  substituted  for  x  throughout  any  equation, 
the  locus  is  translated  a  distance  a  in  the  a;-direction. 

VII.  If  (y  —  b)  be  substituted  for  y  in  any  equation,  the  locus 
is  translated  the  distance  b  in  the  ^/-direction. 

VIII.  The  addition  of  the  term  mx  to  the  right  side  of 
y  =  fi^)  shears  the  locus  y  =  f(x)  in  the  line  y  =  mx. 

IX.  li  {6  —  a)  he  substituted  for  9  throughout  the  polar  equa- 
tion of  any  locus,  the  curve  is  rotated  about  the  pole  through  the 
angle  a. 

X.  If  the  equation  of  any  locus  is  given  in  rectangular  coordi- 
nates, the  curve  is  rotated  through  the  positive  angle  a  by  the 
substitutions 

x  cos  a  +  y  sin  a     for  x 
and 

y  cos  a  —  x  sin  a     for  y. 

Exercises 
1.  Translate  the  curves 

(a)  y  =  2x2,  (e)   y  =  e",  (t)    y  =  -' 


(b)  y     =     -    3X2,  (_/■)     y     =    ^3^  (y)     y     =    _. 

(c)  y  =  log  X,  (g)   y  =  sin  x,  (/b)  y  =  x^, 

2 

(d)  y  =  e-^,  (h)  y  =  cos  x,  (l)    y  =  x^, 

two  units  to  the  right;  three  units  to  the  left;  five  units  up;  one  unit 
down;  two  units  to  the  left  and  one  unit  down.  Sketch  each  curve 
in  its  original  and  translated  position  on  a  sheet  of  squared  paper. 

2.  Shear  each  curve  given  in  Exercise  1  in  the  line  2/  =  ix; 
y  =  —  \x',  y  =  x;  y  =  —  x.  Sketch  each  curve  in  its  original  and 
shearetl  position. 

3.  Write  the  equation  of  each  curve  given  in  Exercise  1  when  re- 
flected in  the  X-axis;  in  the  F-axis;  in  the  line  y  =  x]  in  the  line 
y  =  —  X.     Sketch  each  curve  before  and  after  reflection. 

4.  Rotate  tlie  ^curves 


12  CALCULUS  [§12 

(a)  p  =  a  sin  9,  (e)   p  =  a(l  +  cos  $), 

(b)  p  =  a  cos  6,  (f)   p  =  a(l  —  sin  6), 

(c)  p  =  a  cos  ^  +  6  sin  ^,  (^)  p  =  a(l  +  sin  0), 

(d)  p  =  a  (1  —  cos  6),  (h)  p  =  ad, 

about  the  pole  through  an  angle  o'  7-'  k'  ''^j  ~  o'     *^ketch  each  curve 

in  its  original  and  rotated  position. 

6.  Sketch  the  following  pairs  of  curves  on  squared  paper: 

(a)  y  =  x^  and  y  =  x^  -{-  x. 

(6)  y  =  x^  and  y  =  (x  —  Sy  +  2. 

(c)  y  =  x^  and  y  =  —  a;^  —  2x. 

(d)  2/  =  X*  and  y  =  a;*  —  4a;'  +  Gx'^  —  4a;. 

(e)  y  =  —  2x'^  and  j/  =  fx^. 

(f)  y  =  x^  and  2/  =  |x^. 
(?)  2/  =  sin  X  and  y  =  sin  2x. 
(h)  y  =  sin  x  and  y  =  2  sin  x. 

(i)    y  =  cos  X       and     y  —  sin  (  „  —  x  j . 

6.  Rotate  the  following  curves  about  the  origin  through  the  angle 
indicated. 


(o)  x*  -  J/2  =  a" 

through 

45°. 

(6)  x2  -  y2  =  a2 

through 

-45°. 

(c)    x2  -  7/2  =  a2 

through 

90°. 

(d)  a;2  -  ?/  =  a2 

through 

-90°. 

(e)   a;2  +  2/2  =  a 

through 

a. 

if)   y  =  mx^ 

through 

a. 

CHAPTER  I 
DERIVATIVE 

In  Elementary  Analysis  the  student  investigated  the  dependence 
of  a  function  upon  one  or  more  variables  with  the  help  of  algebra 
and  geometry. 

He  is  now  to  study  a  very  powerful  method  of  investigating  the 
behavior  of  functions,  the  method  of  the  infinitesimal  calculus, 
which  was  discovered  by  Newton  and  Leibnitz  in  the  latter  part 
of  the  17th  century.  This  method  has  made  possible  the  great 
development  of  mathematical  analysis  and  of  its  applications  to 
problems  in  almost  every  field  of  science,  particularly  in  engi- 
neering and  physics. 

13.  Increments.  Let  us  consider  the  following  examples  which 
illustrate  the  principles  of  the  calculus: 

Example  1.  A  steel  bar,  subjected  to  a  tension,  will  stretch, 
and  the  amount  of  stretching,  or  the  elongation,  will  continue  to 
increase  as  the  intensity  of  the  force  applied  increases,  until 
rupture  occurs.  The  elongation  is  a  function  of  the  applied  force. 
In  fact,  if  the  force  is  not  too  great,  so  that  the  elastic  limit  is  not 
exceeded,  experiment  has  shown  that  the  elongation  is  propor- 
tional to  the  applied  force  (Hooke's  Law).  If  we  denote  the 
elongation  by  y  and  the  force  by  x,  the  functional  relation  between 
them  will  be  expressed  by  the  simple  equation 

y  =  kx, 

where  A;  is  a  constant.    This  relation  is  represented  graphically 
by  a  straight  line  through  the  origin,  Fig.  10. 

Suppose  that  after  the  bar  has  been  stretched  to  a  certain  length, 
the  force  is  changed.  This  change  in  the  force  produces  a  cor- 
responding change  in  the  elongation,  an  increase  if  the  force  is 
increased,  a  decrease  if  the  force  is  decreased.  Evidently,  from 
the  law  connecting  the  elongation  and  the  force,  this  change  in  the 

13 


14 


CALCULUS 


[§13 


elongation  is  directly  proportional  to  the  change  in  the  force.  We 
shall  call  the  change  in  the  force  x,  the  increment  of  the  Jorce,  or 
the  increment  of  x,  and  shall  denote  it  by  the  symbol  A.x  (read  "in- 
crement of  x"  or  "delta  x").  The  corresponding  change  in  the 
elongation  we  call  the  increment  of  the  elongation,  or  the  increment 
of  y,  and  denote  it  by  Ly. 

In  Fig.  10  let  P  be  any  point  on  the  line  y  —  kx.    If  x  takes  on 
■an  increment  Ax,  y  takes  on  an  increment  Ay.    We  see  that  the 

Ay  . 
ratio  of  these  mcrements,  i.e.,  the  quotient  -r-  is  entirely  inde- 
pendent of  the  magnitude  and  sense  of  Ax  and  of  the  position  of  P 
on  the  line.     Indeed  this  ratio  is  the  slope  of  the  line.     Here  the 
increment  of  y  is  everywhere  k  times  the  increment  of  x. 


Fig.  10. 


The  relation  between  Ay  and  Ax  can  be  shown  without  the  use 
of  the  figure  as  follows:  If  x  is  given  the  increment  Ax,  y  takes  on 
an  increment  Ay  so  that 

y  +  Ay  =  k{x  +  Ax). 


On  subtracting 

y  =  kx, 

Ay  =  kAx. 

Hence 

Ax         ' 

a  quantity  independent  of  x  and  of  Ax. 

Example  2.     A  train  is  moving  along  a  straight  track  with  a 
constant  velocity,  i.e.,  it  passes  over  equal  distances  in  equal  inter- 


§13] 


DERIVATIVE 


15 


vals  of  time.  Denoting  by  s  the  distance  measured,  say  in  miles, 
from  a  fixed  point,  and  by  t  the  time  measured,  say  in  hours, 
which  has  elapsed  since  the  train  passed  this  point,  the  functional 
relation  between  s  and  t  is  expressed  by 

s  =  ct, 

where  c  is  a  constant  denoting  the  velocity  of  the  train.  This 
function  is  represented  graphically  by  a  straight  line.  Fig,  11.  If 
we  take  an  increment  of  time  At  following  an  instant  t  and  measure 

As 
the  distance  As  passed  over  in  this  time,  the  quotient   r-.  repre- 
sents the  velocity  of  the  train,  since  we  have  assumed  the  velocity 

As 
of  the  train  to  be  uniform.     Furthermore,  the  quotient  -r^  will 


oo 

>^ 

i 

5 

^^^t 

AS 

^ 

o 

Time  ( t ) 

Fig.  11. 

be  independent  of  the  length  of  A^  and  of  the  time  t  to  which  the 
increment  was  given.  As  is  everywhere  c  times  A^.  This  is 
evident  from  the  graph. 

In  these  two  examples  the  functions  were  both  linear  functions 
of  the  independent  variable.  We  have  seen  in  these  cases  (and 
clearly  the  same  is  true  for  any  linear  function,  y  =  ax  -\-  h) 
that  the  ratio  of  A?/  to  Ax  is  constant.  Ay  is  everywhere  equal  to  a 
constant  times  Aa;,  no  matter  how  large  Ax  is  taken  and  no  matter 
at  what  point  (x,  y)  on  the  graph  the  ratio  is  computed. 

Example  3.  Let  us  now  take  an  example  in  which  the  func- 
tional relation  is  no  longer  a  linear  one.  We  shall  find  that  the 
ratio  of  the  increment  of  the  function  to  the  increment  of  the 
variable  is  no  longer  constant.     Suppose  that  the  train  of  Example 

As 
2  is  not  moving  with  constant  velocity.     Then  the  quotient  -r--. 


16 


CALCULUS 


[§M 


IS  called  the  average  velocity  of  the  train  during  the  interval  of 
time  At.  Evidently  this  quotient  will  approximate  more  and 
more  closely  to  a  fixed  value  the  smaller  the  interval  of  time 

At,  is  chosen.     The  limiting  value 

As 
of  the  quotient  -rr  as  At  approaches 

zero  is  called  the  velocity  at  the  time  t. 
Let  the  curve  of  Fig.  12  repre- 
sent graphically  the  relation  be- 
As 
—  tween  s  and  t.    The  ratio  t-;    cal- 


Tlme  (t) 

Fia.  12. 


At 


culated  at  any  point  P  on  the  curve 
is  no  longer  constant  as  in  Example 
2,  but  varies  with  At  and  also  with  the  position  of  the  point  P. 


the     power    function 


14.  The  Fxinction  y  =  x^.  Consider 
y  =  x^.  Let  us  find  the  ratio  of  Ay  to 
Ax  at  a  certain  point  of  the  curve,  say 
(0.2,  0.04),  for  different  values  of  Ax. 
The  results  are  given  in  the  adjoining 
table. 

We  observe  that  as  Ax  is  taken  smaller 

Ay 
and  smaller  the  ratio  v^  approaches  more 

and  more  closely  a  value  in  the  vicinity 
of  0.4. 

Ay 
The  value  of  -r—  will  now  be  calculated 
Ax 

for  any  point  P,  (x,  y),  on  the  curve  y  =x^. 

From    this  value,   which   is   a   function 

of  X  and  Ax,   the  limiting  value  as  Ax  approaches  zero  will  be 

found.    The  point  P,  Fig.  13,  has  the  abscissa  x.     If  we  give  to  x 

an  increment  Ax,  we  have  corresponding  to  the  abscissa,  x  +  Ax, 

the  point  Q  on  the  curve.     Its  ordinate  is 

y  +  Ay  •=  {x  -\-  Ax)2. 

Ay  is  equal  to  the  difference  between  the  ordinates  of  P  and  Q,  or 

Ay—  (x  +  Ax)^  —  x^ 
=  2x  Ax  +  (Ax)2. 


Ax 

Ay 

Ay 

Ax 

0.4 

0.32 

0.8 

0.2 

0.12 

0.6 

0.1 

0.05 

0.5 

0.05 

0.0225 

0.45 

0.02 

0.0084 

0.42 

0.01 

0.0041 

0.41 

0.005  0.002025  0.405 

0.002  0.000804  0.402 

0.0010.0004010.401 

\                 1 

§15] 
Then 


DERIVATIVE 


17 


Ay 
Ax 


=  2x  +  Ax. 


As  Ax  approaches  zero  the  first  term  jn  the  second  member  remains 

unchanged,  while  the  second  term  approaches  zero.     It  follows 

•    •  »  ^y 

that  the  limiting  value  of  v-  as  Ax  approaches  zero  is  2x.     This 

result  is  expressed  by  the  equation 


I 


Fig.  13. 


Ay 


(read  "limit  of  -r^  as  Ax  approaches  zero").     When  x  =  0.2, 


limAy 


=  0.4.  This  is  the  limiting  value  which  the  ratio  tabu- 
lated in  the  last  column  of  the  table  above  is  approaching.  When 
^  =  3,    l^o^x  =  6.     When    x  =  |,    H^.f^  =  1.     Thus  the 

formula  just  obtained  enables  us  to  calculate  very  easily,  for  any 

Ay 
value  of  X,  the  limit  of  -r-  as  Ax  approaches  zero. 

16.  Slope  of  the  Tangent. — The  curve  of  Fig.  14  is  the  graph  of 
the  function  y  =  f(x).     On  this  curve  take  the  point  P  with 


18 


CALCULUS 


[§16 


coordinates  x  and  y,  and  a  second  point  Q  with  coordinates 
X  -\-  Ax  and  y  +  Ay.  Draw  the  secant  PQ  making  tlie  angle  <^ 
with  the  X-axis  and  the  tangent  PT'  making  the  angle  t  with  the 

Ay 
X-axis.     From  the  figure,  -r-  is  the  slope  of  PQ,  or 


Ay 
Ax 


=  tan  (/>. 


As  Ax  is  taken  smaller  and  smaller  the  secant  PQ  revolves  about 
the  point  P,  approaching  more  and  more  closely  as  its  limiting 
position  the  tangent  PT',  and  tan  <^  approaches  tan  t.     (The 


Y 

/  , 

T 
/ 

J    / 

P 

/ 

n 

X 

O 

4 

Fig.  14. 

student  will  recall  that  the  tangent  to  a  curve  at  a  point  P  is 
defined  as  the  limiting  position  of  the  secant  PQ  as  the  point  Q 
approaches  P.)     Hence 


hm  Ay 


aj=oAx       ^^^=0        ^ 


Hence  ^x^qX  ^^  equal  to  the  slope  of  the  tangent  to  the  curve 

y  =  /(^)  oi  the  point  for  which  this  limit  is  computed. 

In  the  case  of  the  parabola,  y  =  x^,  the  slope  of  the  tangent  at 

2 


§15] 


DERIVATIVE 


19 


the  point  (x,  y)  is  2x.  This  shows  that  the  curve  becomes  steeper 
and  steeper  for  larger  positive  and  negative  values  of  x  and  that  at 
X  =  0  the  slope  is  zero. 

In  Fig.  15,  let  the  X-axis  be  divided  uniformly  and  let  the  F-axis 
be  divided  in  such  a  way  that  distances  measured  from  0  on  a 
uniform  scale  are  equal  to  the  squares  of  the  numbers  affixed  to 
the  points  of  division.  Draw 
lines  parallel  to  the  F-axis 
through  equidistant  points 
on  the  X-axis  and  lines  par- 
allel to  the  X-axis  through 
points  on  the  F-axis  whose 
affixed  numbers  on  the  non- 
uniform scale  are  equal  to 
the  numbers  affixed  to  the 
points  on  the  X-axis  through 
which  lines  were  drawn. 

On  the  cross  section  paper 
thus  constructed,  any  point 
at  the  intersection  of  a  hori- 
zontal and  a  vertical  line 
bearing  the  same  number  is 
a  point  on  the  curve  ij  =  x^ 
which  would  be  constructed 
in  the  usual  way  by  using 
the  uniform  scale  on  the  F- 
axis  as  well  as  on  the  X-axis. 

Join  the  consecutive  points 
thus  located  by  straight  lines. 
These  lines  are  the  diagonals 
of  the  rectangles  on  the  cross 
section  paper  and  they  are 
secants  of  the  parabola  y  =  x 

let  PR  =  ^x 


Y 

2 

1 

/ 

8 

Q 

J 

it 

p 

1 

R 

Y 

1 

3 

/ 

4 
1 

/ 

2 

1 

y 

4 

_^,,„0^^ 

„ 

0 

~ 

I  - 

^ 

' 

.     . 

f      - 

r      ; 

[ 

Fig.  15. 

Let  PQ  be  such  a  diagonal  and 
RQ      Ay 


Then  RQ  =  Ay  and  ^  =  ^.  the   slope  of  the 

secant  PQ.  The  diagonals  give  an  approximate  idea  of  the  slope 
of  the  curve.  The  construction  shows  why  the  slope  increases 
so  rapidly  with  x. 


20  CALCULUS  [§16 

As  more  and  more  horizontal  and  vertical  lines  are  inserted,  the 
diagonals  approach  more  and  more  nearly  the  direction  of  the 
tangent  lines. 

The  fact  that  the  slope  of  the  tangent  to  the  parabola  y  =  x^ 
is  2x  furnishes  an  easy  way  of  constructing  the  tangent  at  any 
point  P  {x,  y).  We  have  only  to  draw  from  P  in  the  direction  of 
the  positive  X-axis,  a  line  PK  of  unit  length,  and  from  the  ex- 
tremity of  this  line,  a  line  KT  parallel  to  the  F-axis,  whose  length 
is  twice  the  abscissa  of  P.  The  line  joining  PT"  is  the  tangent  to  the 
parabola  at  P.  When  the  abscissa  is  negative  the  line  XT  is  to  be 
drawn  downward. 

16.  Maxima   and    Minima.     The    algebraic   sign    of  ^^q  ~ 

enables  us  to  tell  at  once  where  the  function  y  is  increasing  and 
where  it  is  decreasing  as  x  increases.  For,  if  the  slope  is  positive 
at  a  point,  the  function  is  increasing  with  x  at  that  point  and  the 
greater  the  slope  the  greater  the  rate  of  increase.  Similarly  if  the 
slope  is  negative,  the  function  is  decreasing  as  x  increases.  Hence 
the  function  y  =  x^  is  a  decreasing  function  when  a;<0  and  an 
increasing  function  when  x>0,  since  the  slope  is  equal  to  2x. 
When  X  =  0  the  slope  is  zero  and  the  tangent  is  parallel  to  the 
X-axis.  Since  the  function  is  decreasing  to  the  left  of  x  =  0 
and  increasing  to  the  right  of  this  line,  it  follows  that  the  function 
decreases  to  the  value  zero  when  x  =  0  and  then  increases.  This 
value  zero  is  a  minimum  value  of  the  function  y  =  x"^.  In  general 
we  define  minimum  and  maximum  values  of  a  function  as  follows: 

Definition.  Let  y  =  f{x),  where /(x)  is  any  function  of  a  single 
argument.  If  y  decreases  to  a  value  m  as  x  increases  and  then 
begins  to  increase,  m  is  called  a  minimum  value  of  the  function.  If 
y  increases  to  the  value  M  as  x  increases  and  then  begins  to  decrease, 
M  is  callec  a  maximum  value  of  the  function. 

Thus  in  Fig.  16,  if  ABDFHI  is  the  graph  oi  y  =  f(x),  the  func- 
tion increases  to  the  value  represented  by  the  ordinate  bB  and 
then  begins  to  decrease.  bB  is  then  a  maximum  value  of  the 
function.  Similarly  fF  is  another  maximum  value.  dD  and  hH 
are  minimum  values  of  the  function. 

In  referring  to  the  graph  of  a  function,  points  corresponding  to 
maximum  and  points  corresponding  to  minimum  values  of  the 


§17] 


DERIVATIVE 


21 


function  will  be  called,  respectively,  the  maximum  and  minimum 
points  of  the  curve.  Thus  B  and  F,  Fig.  16,  are  maximum  points 
and  D  and  H  are  minimum  points  of  the  curve. 

Thus,  zero  is  a  minimum  value  oiy  ==  x^  or  (0,  0)  is  a  minimum 
point  on  the  curve  y  =  x^. 

It  will  be  noticed  that  a  maximum  value,  as  here  defined,  is  not 
necessarily  the  largest  value  of  the  function,  nor  is  a  minimum 
value  the  smallest  value  of  the  function.  A  maximum  value  may 
even  be  less  than  a  minimum  value. 


17.  Derivative.  We  see  that  the  limit  of  the  ratio  of  the  incre- 
ment of  the  function  to  the  increment  of  the  independent  variable 
as  the  latter  increment  approaches  zero,  is  very  useful  in  studying 
the  behavior  of  the  function.  This  limit  is  called  the  derivative  of 
the  function  with  respect  to  the  variable.  Hence  the  following 
definition: 

The  derivative  of  a  function  of  a  single  independent  variable 
with  respect  to  that  variable  is  the  limit  of  the  ratio  of  the  increment 
of  the  function  to  the  increment  of  the  variable  as  the  latter  increment 
approaches  zero.    The  derivative  of  a  function  y  with  respect  to  a 

dy 
variable  x  is  denoted  by  the  symbol  ^'    This  symbol  will  not  be 

considered  at  present  as  representing  the  quotient  of  two  quan- 
tities but  as  a  symbol  for  a  single  quantity.  Later  it  will  be 
interpreted  as  a  quotient.  (See  §G1.)  It  is  read,  "the  derivative 
of  y  with  respect  to  x."  The  process  of  finding  the  derivative 
is  called  differentiation. 


22 


CALCULUS 


[§18 


18.  Velocity  of  a  Falling  Body.  As  a  further  illustration  of  the 
application  of  the  derivative  let  us  attempt  to  find  the  velocity  of 
a  falling  body  at  any  instant.  The  law  of  motion  has  been  experi- 
mentally determined  to  be 

s  =  yt\ 

where  s  is  the  distance  through  which  the  body  falls  from  rest  in 
time  t.  If  s  is  measured  in  feet  and  t  in 
seconds,  the  constant  g  is  32.2  feet  per 
second  per  second,  s  is  plotted  as  a  func- 
tion of  the  time  in  Fig.  17.  At  any  time 
t,  let  t  take  on  an  increment  A^  s  will  take 
on  an  increment  As,  represented  in  the 
figure  by  the  line  RQ.    Since  s  =  \gt^, 


1 

•0 

Ql 

o 

» 

a 

S 

AS 

P 

P/    X 

R 

Ac^t 

Time  (t) 

Fig.  17. 


Hence 


s  +  As  =  ^git  +  Aty. 

As  =  hgit  +  Aty  -  hgt\ 


(1) 


or 


As  =  gtAt  +  hg{Aty. 


(2) 

This  is  the  distance  through  which  the  body  falls  in  the  interval 

As 
At  counted  from  the  time  t.     The  quotient  -r-  is  the  average 

velocity  for  the  interval  At.     The  velocity  at  t  has  been  de- 

1"     As 
fined  as  ^™  ^^,  i.e.,  as  the  derivative  of  s  with  respect  to  t. 

To  find  this  limit  divide  (2)  by  At  and  obtain 
As 


At 


=  gt  +  \gAt, 


the  average  velocity  for  the  interval  A^.     From  which 

lira  ^ 


At  =  0  At 


=  gt, 


or 


ds 
dt 


=  gi, 


(3) 


(4) 


the  velocity  at  t.  Thus  the  velocity  at  the  end  of  three  seconds 
is  96.6  feet  per  second;  at  the  end  of  four  seconds,  128.8  feet  per 
second. 


§20]  DERIVATIVE  23 

19.  Illustration.  As  an  example  of  the  use  of  the  derivative  in 
studying  the  behavior  of  a  function,  let  us  consider  the  power 
function 

y  =  x^. 
y  +  Ay  =  (x  +  Axy, 

y  +  Ay  =  x^  -\-  3a;2(Ax)  +  3x(Aj;)2  +  {Axy, 
Ay  =  3x2(Aa;)  +  3x(Aa;)2  +  (Ax)^, 


^  =  3x2  +  SxiAx)  +  (Ax)\ 


Then 


Hm  Ay  _ 
Ax=0  Ax~ 

T-  =  3x2. 
ax 

For  X  =  0  the  derivative  is  equal  to  zero  and  consequently  the 
tangent  at  (0,  0)  is  horizontal  and  coincides  with  the  X-axis. 
For  all  other  values  of  x  the  derivative  is  positive.  This 
shows  that  the  function  is  an  increasing  function  for  all  these 
values  of  x.  Where  is  the  slope  of  the  curve  equal  to  1? 
Equal  to  Vs? 

20.  Illustration.    The  solution  of  the  following  problem  will 
further  illustrate  the  use  of  the  derivative. 

Find  the  dimensions  of  the  gutter  with 

the    greatest   possible   carrying   capacity  ^ ^ p 

and  with  rectangular  cross  section,  which 

can  be  made  from  strips  of  tin  30  inches 

wide  by  bending  up  the  edges  to  form  ^ 

the  sides.     See  Fig.  18.  Fig.  18. 

If  the  depth  MR  is  determined,   the 
width  is  also  determined,  since  the  sum  of  the  three  sides  MR, 
PQ,  and  RQ  is  30  inches.     We  seek  to  express  the  area  of  the 
cross  section  as  a  function  of  the  depth.     Denote  the  depth  by 
X  and  the  area  by  A.     The  width  RQ  is  30  —  2x.     Hence 

A  =  (30  -  2x)x. 
In  Fig.  19,  A  is  plotted  as  a  function  of  x.     A  first  increases  with  x 


24 


CALCULUS 


[§20 


and  then  decreases.  The  value  of  x  for  which  A  reaches  its 
greatest  value  can  be  determined  from  a  graph  with  a  high  degree 
of  approximation.  The  derivative  can  be  used  to  calculate 
accurately  this  value  of  x  and  this  saves  construction  of  an 
accurate  graph. 

From  0  to  H,  A  is  an  increasing  function.  Its  derivative  is 
therefore  positive  for  this  part  of  the  curve .  From  H  to  N  the 
function  A  is  decreasing.  Its  derivative  is  therefore  negative 
for  this  part  of  the  curve.  At  the  point  H  the  derivative  changes 
sign,  passing  from  positive  values  through  zero  to  negative  values. 
The  abscissa  of  the  point  H  can  then  be  found  by  finding  the 


Fig.  19. 


derivative  of  A  with  respect  to  x  and  determining  where  it  changes 
sign.  In  this  case  the  change  of  sign  occurs  where  the  derivative 
is  equal  to  zero.     We  find  by  the  method  of  increments 


ax 


Ax  =  4(7.5  -  x). 


=  0  when  x  =  7.5.     If  a:<7.5,  ~r~  is  positive  and  A  is  an 


dx 


dx 


dA 


increasing  function.  If  x>7.5,  -7—  is  negative  and  A  is  a  de- 
creasing function.  This  shows  that  A  increases  up  to  a  certain 
value  at  a;  =  7.5  and  then  begins  to  decrease.     Hence  the  gutter 


§20] 


DERIVATIVE 


25 


will  have  the  greatest  cross  section  if  its  depth   be  made  7.5 
inches. 

It  is  interesting  to  plot  the  derivative  as  a  function  of  x  on  the 
same  axes.  See  the  dotted  line,  Fig.  19.  The  statements  made 
concerning  the  derivative  are  verified  in  the  graph. 


Exercises 

1.  Consider  the  function  y  =  /(x)  whose  graph  is  given  in  Fig.  20. 
In  what  portions  of  the  curve  is  the  derivative  positive?  In  what 
portions  negative?    Where  is  the  derivative  equal  to  zero? 


y  =f(x) 


Fig.  20. 


2.  Find  ,,  U  y  =  3x*.  For  what  values  of  x  is  the  function  in- 
creasing? For  what  values  decreasing?  At  what  point  does  the 
tangent  line  drawn  to  the  curve  representing  the  function,  make  an 
angle  of  45°  with  the  positive  direction  of  the  axis  of  x?  Find  the 
coordinates  of  the  maximum  or  minimum  points  on  the  curve. 

3.  Answer  questions  asked  in  Exercise  2,  if  y  =  x'. 

4.  Answer  questions  asked  in  Exercise  2,  ii  y  =  x*. 

5.  Answer  questions  asked  in  Exercise  2,  if  y  =  x^. 

6.  Answer  questions  asked  in  Exercise  2,  if  y  =  x*  —  2x  +  3. 

7.  Answer  questions  asked  in  Exercise  2,  ii  y  =  -^ 2"  +  2x  —  6. 

8.  Find  the  derivative  of  \/x. 
Solution.     Let  y  =  \/x. 

Then 

y  +Ay  =  y/x  +Ax 
and 

Ay  =  y/x  +  Ax  —  Vx 
Ay  _  \^x  +  Ax  —  \/x 
Ax  ~  Ax 


26  CALCULUS  [§20 

Rationalize  the  numerator : 

Ay  1 


Ax       Vx  +  Ax  +  Vx 

As  Ax  approaches  zero  the  right-hand  side  of  this  equation  approaches 

-7^- 7^-      Then 

Vx  +  vx 


or 


lira   A^ 

1 
2Vx 

1 

2\/x 

9. 

Find  the  derivative  of  v/x  - 

^. 

lU. 

Find  the  derivative  of  \/x*  - 

-4. 

CHAPTER  II 
LIMITS 

In  §17  the  derivative  was  defined  as  the  limit  of  a  certain  ratio. 
The  word  limit  was  used  without  giving  its  precise  definition, 
as  the  reader  was  supposed  to  have  a  fair  conception  of  the  mean- 
ing of  this  term  from  previous  courses  in  mathematics.  How- 
ever, since  the  entire  subject  of  the  calculus  is  based  on  limit 
processes  it  is  well  to  review  the  precise  definition  and  to  state 
certain  theorems  from  the  theory  of  limits. 

21.  Definition.  //  a  variable  changes  by  an  unlimited  number 
of  steps  in  such  a  way  that,  after  a  sufficiently  large  number  of  steps, 
the  numerical  value  of  the  difference  between  the  variable  and  a 
constant  becomes  and  remains,  for  all  subsequent  steps,  less  than  any 
preassigned  positive  constant,  however  small,  the  variable  is  said  to 
approach  the  constant  as  a  limit,  and  the  constant  is  called  the  limit 
of  the  variable. 


1 


K 


A  Xi  X2         Xs  Xi  B 

Fig.  21. 

Illustration  1.  Let  AB,  Fig.  21,  be  a  line  two  units  in  length, 
and  let  x  be  the  distance  from  A  to  a  point  on  this  line.  Suppose 
that  X  increases  from  0  by  steps  such  that  any  value  of  x  is  greater 
than  the  preceding  value  by  one-half  of  the  difference  between  2  and 

2  —  X 
this  preceding  value,  i.e.,  by  — ^ —     Xi,  Xi,  Xz,  Xi,    •    •    -   are  the 

end  points  of  the  portions  of  the  line  representing  the  successive 
values  of  x.  Then  the  lengths  X\B  =  1,  x^B  =  ^,  XzB  =  (^)^ 
XiB  =  (^)',  •  •  •  ,  XuB  =  (^)""^  are  the  successive  differ- 
ences between  the  constant  length  2  and  the  variable  length  x. 
This  difference  becomes  and  remains  less  than  any  preassigned 
length  KB  after  a  sufficient  number  of  steps  has  been  taken. 

27 


28 


CALCULUS 


[§21 


This  is  true  however  small  the  length  KB  is  chosen.  Therefore, 
by  the  definition  of  the  limit  of  a  variable,  2  is  the  limit  of  the 
variable  x. 

Illustration  2.     Consider  the  variable  x"^  —  2.     Give  to  x  the 

3(2"  -  1) 


vnliiAQ  ft     ^     A     iJ-     A.1     A3. 

values  u,  2 ,  4 ,    a  i    1 « ,   32, 


2" 


=  3[x-i]. 


which  are  chosen  by  starting  with  x  =  0  and  giving  to  it  successive 
increments  which  are  one-half  the  difference  between  3  and  the 


7 

B 

8 

6 

R 

0 

5 

a 

4 

«3 

^ 

, 

2 

< 

) 

0 

1 

a 

! 

5 

i 

\ 

Fig.  22. 

preceding  value  of  x.  The  corresponding  values  of  a;^  —  2  are 
given  in  the  adjoining  table.  The  corresponding  points,  excepting 
(0,  -  2),  are  plotted  in  Fig.  22. 

From  the  table  and  the  expression  a;^  —  2  it 
is  readily  seen  that  the  difference  between  7 
and  the  variable  x^  —  2  becomes  and  remains 
less  than  any  previously  assigned  quantity  (such 
as  KB,  Fig.  22)  after  a  sufficiently  large  number 
of  steps.  Therefore  7  is  the  limit  of  the  vari- 
able a;2  —  2  as  X  approaches  3. 

Illustration  3,     By  giving  x  values   nearer 

and  nearer  2,  the  value  of ?:  becomes  nu- 

'  x  —  2 

merically  larger  and  larger.     Indeed  its  numeri- 
cal value  can  be  made  greater  than  any  preas- 
signed  positive  number  however  large  by  choosing  x  sufficiently 


X 

x''  -2 

0.0 

-2.00 

1.5 

0.25 

2.25 

3.06 

2.62 

4.89 

2.81 

5.91 

2.91 

6.45 

2.95 

6.72 

2.98 

6.86 

2.99 

6.93 

§23]  LIMITS  29 

near  2.     The  variable does  not  approach  a  limit  as  x  ap- 

proaches  2.     Instead  of  doing  so  it  increases  without  limit. 

//  a  variable  changes  by  an  unlimited  number  of  steps  in  such 
a  way  that  after  a  sufficiently  large  number  of  steps  its  numerical 
value  becomes  and  remains,  for  all  subsequent  steps,  greater  than 
any  preassigned  positive  number  however  large,  the  variable  is  said 
to  become  infinite.  Illustration  3  of  this  section  is  an  example 
of  a  variable  which  becomes  infinite,  or  approaches  infinity. 

22.  Notation.  If  in  any  limit  process,  the  variable,  say  y,  is  a 
function  of  another  variable,  say  x,  the  successive  steps  by 
which  y  changes  are  determined  by  those  by  which  x  changes. 
//  y  approaches  a  limit  A,  as  x  approaches  a  limit  a,  we  say  that  the 
limit  of  y  as  X  approaches  a  is  A,  and  write 

lim .,  _ 


x=a 


y  =  A. 


After  what  has  just  been  said,  the  meaning  of  the  two  following 
expressions  will  be  clear: 

lim  y  ^  ^  lim„  =  ^o. 

In  the  second  case  a  limit  does  not  really  exist.  The  form  of 
expression  is  only  a  convenient  way  of  saying  that  if  x  is  taken 
sufficiently  near  a,  the  value  of  y  can  be  made  to  become  and 
remain  greater  in  numerical  value  than  any  preassigned  positive 
number  however  large. 

From  the  illustrations  of  the  preceding  section  we  have: 

1.  ""^  X  =  2,  where  n  is  the  number  of  steps  taken. 

2.  ^i"^ 

z=3 

Q    lim 


X.3  (^^  -  2) 


"2=   "'^ 


23.  Infinitesimal.    In  the  particular  case  where  the  limit  of  a 

variable  is  zero,  the  variable  is  said  to  be  an  infinitesimal.  An 
infinitesimal  is  a  variable  whose  limit  is  zero.  Thus  Ay  and  Ax 
which  were  used  in  §§13,  14,  and  15  are  thought  of  as  approaching 


30  CALCULUS  [§24 

zero  and  are  infinitesimals.  Hence  the  derivative,  §17,  is  defined 
as  the  limit  of  the  quotient  of  two  infinitesimals.  Infinitesimals 
are  of  fundamental  importance  in  the  Calculus.  Indeed  the 
subject  is  often  called  the  Infinitesimal  Calculus. 

24.  Theorems  on  Limits.  The  following  theorems  concerning 
limits  are  stated  without  proof: 

Theorem  I.  7/  two  variables  are  always  equal  and  if  one  approaches 
a  limit,  the  other  approaches  the  same  limit. 

Theorem  II.  The  limit  of  the  sum  of  tivo  variables,  each  of  which 
approaches  a  limit,  is  equal  to  the  sum  of  their  limits. 

Theorem  III.  The  limit  of  the  difference  of  two  variables,  each 
of  which  approaches  a  limit,  is  equal  to  the  difference  of  their  limits. 

Theorem  IV.  The  limit  of  the  product  of  two  variables,  each  of 
which  approaches  a  limit,  is  equal  to  the  product  of  their  limits. 

Theorem  V.  The  limit  of  the  quotient  of  two  variables,  each  of 
which  approaches  a  limit,  is  equal  to  the  quotient  of  their  limits, 
provided  the  limit  of  the  divisor  is  not  zero. 

If  the  limit  of  the  divisor  is  zero,  the  quotient  of  the  limits  in 
Theorem  V  has  no  meaning,  since  division  by  zero  is  an  impossible 
operation.  For,  the  quotient  Q  of  two  numbers  A  and  B  is 
defined  as  the  number  such  that  when  it  is  multiplied  by  the 
divisor  B,  the  product  is  the  dividend  A.  Now  if  B  is  zero  while  A 
is  not  zero,  there  clearly  is  no  such  number. 

25.  The  Indeterminate  Form  5.    If,  in  the  quotient  considered 

above,  A  is  also  zero,  any  number  will  satisfy  the  requirement, 

so  that  Q  is  not  determined.     One  encounters  exactly  this  diffi- 

a;2  _  4 
culty  in  seeking  the  value  of  ^  at  a;  =  2.     Its  value  is  not 

determined  at  this  point  but  it  is  determined  for  all  finite  values 
of  X  different  from  2.  We  define  its  value  at  a;  =  2  as  the  limit  of 
its  value  as  x  approaches  2.  The  student  should  construct  a 
graph  of  this  function.  Usually  we  proceed  as  follows  to  find  the 
desired  limit. 

lim  a;2  -  4        Hm 


x=2  a;  _  2         x=2 


(x  +  2)  =  4. 


a;2    _  4 

The  expression     -_-^  is  said  to  be  indeterminate  at  x  =  2, 


§26] 


LIMITS 


31 


since  any  one  of  an  infinite  number  of  values  can  be  assigned  to  it. 
The  determination  of  its  limiting  value  as  x  approaches  the  value  2 
is  called  the  evaluation  of  the  indeterminate  form.  Indeterminate 
forms  of  this  and  other  types  are  frequently  found  in  the  Calculus. 


Thus  V-  is  an  indeterminate  form  for  Ax 


0.     We  have  already 

seen  in  several  cases  how  it  can  be  evaluated.  Exactly  as  in  the 
example  just  given,  we  have  sought  the  limit  of  the  quotient  as 
Ax  approaches  zero  and  not  the  quotient  when  Ax  =  0,  because  the 
latter  quotient  has  no  meaning. 


Exercises 

1.  Determine  the  following  limits,  if  they  exist. 


(6)       ^^gCOtX. 


,      lim 

(a)        ^cosx. 

(c)  \"}  sin  -.     Draw  the  curve  for  values  of  x  between  —  tt  and  -f-  jr. 

(d)  ".'V,xsin"- 

2.  Evaluate  the  following  indeterminate  forms : 


(a) 


x2  -9 


X  -3 
3.  Find 


(&) 


X*  +  6x2| 


symbol 


z=s=3  '  '   3x3  +  x^\x^O 

lim  3x2 _    jjj^  4^_    jjj^  4^2^    jj^^  4x^  +  3 

j;=oo     a;    »      x=oo53.2>      a;===co5^2>      x=  «=         Qx^       ' 

Show  that  it  is  an  indeterminate  form. 


26,  Continuous    and    Discontinuous    Functions, 
graphs  of  the  following  functions: 

1 


1.  y  = 


2.  y  =  x^. 


3.  y  =  7x  + 


Discuss    the 

Draw     the 
1 


4.  7/  =  tan  X.  5.  y  =  sin  x.  6-  y  =  3^. 

Hint.  In  6,  values  in  the  vicinity  of  a;  =  0  should  be  carefully 
determined.  Take  a  set  of  values  of  x  approaching  0  from  the 
left  and  another  set  approaching  it  from  the  right. 


7.  2/  =  3^ 


3''  + 2 
3'  +  1 


32  CALCULUS  [§26 

Study  the  vicinity  of  a;  =  0.     See  6. 

The  functions  2,  5,  and  7  are  said  to  be  continuous  while 
1,  3,  4,  6,  and  8  are  discontinuous.  The  meaning  of  these 
terms  is  obvious  from  the  graphs  that  have  been  drawn.  A 
precise  definition  follows :    A  function  f{x)  is  said  to  be  continuous 

atx  =  a  if  Ji^  fix)  = /(a). 

In  1,  3,  4,  6,  and  8,  this  condition  is  not  satisfied  at  a;  =  0, 

0,  ^,  0,  and  0,  respectively.    In  these  examples  the  functions 

either  become  infinite  for  the  values  of  x  in  question  or  approach 
different  limits  as  the  value  of  x  is  approached  from  larger  or 
smaller  values.  A  function  f{x)  is  said  to  be  continuous  in  an  in- 
terval (c,  d),  i.e.,  the  interval  c  ^  x  ^  d,  if  it  is  continuous  at  every 
point  in  this  interval.  Thus  the  functions  2,  5,  and  7  are  continu- 
ous in  any  finite  interval.  The  remaining  functions  are  continu- 
ous in  any  interval  not  containing  the  points  to  which  attention 
has  been  called. 


CHAPTER  III 
THE  POWER  FUNCTION 

27.  In  Chapter  I  the  derivative  of  a  function  was  found  by  what 
may  be  called  the  fundamental  method,  viz.,  by  giving  to  the 
independent  variable  an  increment,  calculating  the  corresponding 
increment  of  the  dependent  variable,  and  finding  the  limit  of  the 
ratio  of  these  increments  as  the  increment  of  the  independent 
variable  approaches  zero.  This  method  is  laborious  and  since  it 
will  be  necessary  to  find  derivatives  in  a  large  number  of  problems, 
rules  will  be  established  by  means  of  which  the  derivatives  of 
certain  functions  can  be  written  down  at  once.  The  process  of 
finding  the  derivative  of  a  function  is  called  differentiation. 

In  this  chapter  we  shall  find  the  derivative  of  the  power 
function,  and  study  the  function  by  means  of  this  derivative. 

The  graphs  oi  y  =  x",  for  various  values  of  n,  appear  in  Figs. 
1,  2,  3,  and  4.  If  n  is  positive,  the  curves  go  through  the  points 
(0,  0)  and  (1,  1),  and  are  said  to  be  of  the  parabolic  type.  In  this 
case  a;"  is  an  increasing  function  of  x  in  the  first  quadrant.  If  n 
is  negative,  the  curves  go  through  the  point  (1,  1)  but  do  not  go 
through  the  point  (0,  0).  They  are  asymptotic  to  both  axes  of 
coordinates.  These  curves  are  said  to  be  of  the  hyperbolic  type. 
In  this  case  x^  is  a  decreasing  function  of  x  in  the  first  quadrant. 

The  law  of  the  power  function,  as  stated  in  §3,  should  be 
reviewed  at  this  point. 

28.  Derivative  of  x°.  Let  y  =  x",  where  n  is  at  first  assumed 
to  be  a  positive  integer. 

y  -{-  Ay  =  (x  +  A.r)". 

y+  Ay  =  X-  +nx--''Ax  +  "^^,"~-"^^x"-2  (Ax)^  +     ■     +(Ax)". 

Ay  =  nx--^Ax  +  ^^^  a-"-^  {AxY+     ■  ■  +(Ax)". 
If 
a  33 


34  CALCULUS  [§28 


Ay  ,   ,  n{n  -  1) 


Ax 
lim  Ay 

or 

dy 
dx 


+         .^      a;»-2  Aa;+  •  •  •  +(Ax)»-i. 


=  nx"" 


=  nx"-^  (1) 


This  proof  holds  when  n  is  a  positive  integer.  In  §§33  and  42 
it  will  be  shown  that  the  formula  obtained  holds  for  fractional  and 
negative  exponents.  For  the  present  we  shall  assume  the 
formula  true  for  these  exponents. 

Illustrations. 

1.  ^  =  3.=. 
ax 

dx  dx  X* 

3.  ^'  .  W. 
dt 


dv    ~  ""^      ~  2\/y 


Exercises 

dy .  .  . 

Find  -y-  in  each  of  the  following  fifteen  exercises: 

1.  y  =  x^.  6.  y  =  x.  11.  y  =  x*. 

2.  y  =  x<.  7.  y  =  x*.  12.  y  =  x^. 

3.  y  =  x^  8.  y  =  x».  18.  y  =  x"*. 

3  _1 

4.  y  =  x^".  9.  y  =  x^.  14.  y  =  x   ^. 
6.  y  =  -•                      10.  y  =  -^^  15.  y  =  -,- 

16.  Find  the  slope  of  each  of  the  curves  of  Exercises  1-15  at  the 
point  (1,  1);  also  at  the  point  whose  abscissa  is  ^  and  whose  ordinate 
is  positive. 

17.  By  making  use  of  the  derivative,  find  for  what  values  of  x 
each  function  given  in  Exercises  1-15  is  increasing;  is  decreasing. 


§29]  THE  POWER  FUNCTION  35 

18.  How  does  the  slope  of  j/  =  a;"  change  with  increasing  x,  if  x 
is  positive  and  if  n  is  positive  and  less  than  1?  If  n  is  positive  and 
greater  than  1  ? 

19.  Find  where  the  slope  of  each  curve  given  in  Exercises  1-15  is 
equal  to  zero;  equal  to  1. 


20.  Find:riif: 
at 

(a)  s  =  t\ 

(C)     8    =   Vt' 

(e)  s  =  ^yr^ 

(b)   s=l,- 

(d)  s  =  \- 
t^ 

CO  .  =  ^-- 

21.  Find  ^f  if: 
at 

(a)  y  =  IK 

(c)  y  =  ^l 

(e)  y  =  IK 

ib)y=l 

(d)  y  =  4- 

U)y=l- 

29.  The  Derivative  of  ax°.  In  case  the  power  function  is  writ- 
ten in  the  more  general  form  ax",  it  is  easy  to  see  that  the  con- 
stant multiplier  a  will  appear  as  a  coefficient  in  all  terms  on  the 
right-hand  side  of  the  equations  in  the  proof  in  §28,  and  the 
derivative  oi  y  =  ax"  is,  therefore, 

f^  =  nax'^-^,  (1) 


or 

d(ax'') 
dx 


—  nax"-^.  (2) 


The  proof  of  the  formula  is  for  positive  integral  values  of  n  only, 
but  as  in  §28  will  be  assumed  for  all  commensurable  exponents.^ 

Since  ax"-^  is  the  given  power  function  y  =  ax"  divided  by  x, 
formula  (1)  may  be  written 

'  The  relation  of  formula  (2)  to  that  of  §28  is  at  once  evident  when  it  is  recalled  that 
the  curve  J/  =  oi"  can  be  thought  of  as  obtained  from  the  curve  j/  =  x"  by  stretching 
all  ordinates  in  the  ratio  1 :  a.  Then  the  slope  of  the  tangent  at  a  point  of  j/  =  ox"  is 
o  times  the  slope  of  the  tangent  to  y  =  i"  at  the  corresponding  point;  i.e., 

d(ax")         d(x'^)  , 

=  o =  anx"     . 

dx  dx 


36 


CALCULUS 


[§29 


The  geometrical  meaning  of  formula  (3)  is  shown  by  Fig.  23. 

y 

The  fraction  -  is  the  slope  of  the  radius  vector  OP  from  the  origin 

to  the  point  P  on  the  curve.  Formula  (3)  states  that  the  slope, 
at  any  point  of  the  graph  of  the  function,  y  =  ax",  is  n  times  the 
slope  of  the  radius  vector  OP.  Thus,  if  n  =  1,  ?/  =  ax"  reduces 
to  a  straight  line  through  the  origin,  and  the  line  has  the  same 
slope  as  OP.  If  n  =  2  the  curve  is  the  parabola  y  =  ax^,  and  the 
slope  of  the  curve  is  always  twice  that  of  OP.    If  n  =  —1  the 


r 

F 

ft  1 

/ 

o 

Fig.  23. 

curve  is  the  rectangular  hyperbola,  y  = 

curve  is  the  negative  of  the  slope  of  OP. 
Illustrations. 
d{7x^) 


1. 


dx 

il 

dx 


=  14a;. 


d{5x-^)       ^  d(x-^) 


and  the  slope  of  the 


dx 


=  5 


dx 


=    -lOx-3   =    - 


10 


3.    -^-iT--  =  6  ^TT-  =  9^2 


dt 


4.   If  2/  =  5x', 


dt 


dy 
dx 


§30]  THE  POWER  FUNCTION  37 

5.    If  V  =  ~2'      ^~  =    -2 


x^       dx  X 


Exercises 
Find  -T~  in  each  of  the  following  fifteen  exercises . 

1.  y  =  4x3.  6.  2/  =  2xi  11.  y  =  -3x«. 

2.  J/  =  3\/x.  7.  2/  =  4x^.  12.  ?/  =  x. 

3.  y  =  5x*.  8.  y  =  fx"".  13.  y  =  -x. 

4.  y  =  3x.  9.  2/  =  lOv^i.  14.  2/  =  -3\/x. 

2  4  4 

6.  2/  =    -,'  10.  2/  =  --2-  15.  2/  =  --3- 

16.  Find  -yj  in  each  of  the  following: 

(a)  s  ==  2<2.     (6)  s  =  SVi".     (c)  s  =  -^tK 

civ 

17.  Find  -jt  in  each  of  the  following: 

(a)  2/  =  4v^.     {b)  y  =  -  At.     (c)  ?/  =  -Si^. 

18.  Find  the  slope  of  each  of  the  curves  given  in  Exercises  1-15, 
at  the  point  whose  abscissa  is  1;  at  the  point  whose  abscissa  is  |. 

19.  For  what  values  of  x  is  each  function  given  in  Exercises  1-15 

increasing?     Decreasing?     Where,  if  at  all,  is  the  slope  of  each  of 

these  curves  zero? 

2 

20.  Draw  the  curves  y  =  \x^,  y  =  ~,  y  =  x^,  y  =  -y/x]  and  draw 

tangent  lines  to  them  at  the  points  for  which  the  abscissas  are  1,  2,  3, 
and  4.  Make  a  table  showing  the  slope  of  the  radius  vector  and  the 
tangent  line  for  each  of  these  points. 

30.  Rate  of  Change  of  ax°.  Let  y  =  ax",  where  a;  is  a  function 
of  the  time  t.  Since  a:  is  a  function  of  t,y  is  a,  function  of  t.  For 
example,  y  =  Sx^,  where  x  =  t  —  1. 

Let  Ax  and  Ay  be  the  increments  of  x  and  y,  respectively,  corre- 

Aw 
spending   to   the   increment   A^  of   t.     -rr  is  the  average  rate  of 

dv 
change  of  y  during  the  interval  AL      .r  is  the  rate  of  change  of  y 

at  the  instant  t. 
At  any  time,  t 

y  =  ax". 


38  CALCULUS  [§30 

At  the  time  t  +  Af, 

y  -\-  Ay  =  a{x  +  Ax)". 

y  -\-Ay=  alx-+nx"-'Ax  +         Z    ^x»-''(Ax)H h  (Ax)"l  • 

Ay  =  aLx"-'  +  "^^^^  a;»-2Aa;  +    •   •    •  +  (A.t)"-i1  Ax. 

Aw  r  ,         >       W(W     "~     1)  OA  ■  I         /A       N  iTAX 

^^  =  a   wx»-i  +      ^2"^  x"-2Ax  +    •    •    ■  +  (Ax)»-i  ^• 

As  Af  approaches  zero,  the  expression  within  the  brackets  ap- 
proaches nx""^,  and  irr  approaches  -jr* 

Hence 

dy  dx 


dt  -°^^""'dr 

or 

.     dt     "*^       dt 

The  rate  of  change  of  the  function  ax"  is  expressed  in  terms  of  x 

dx 
and  of  -TT,  the  rate  of  change  of  x.     If  then  the  rate  of  change  of 

X  for  a  given  value  of  x  is  known,  the  rate  of  change  of  the  func- 
tion for  that  value  of  x  can  be  calculated. 

Illustration  1.  The  side  of  a  square  is  increasing  at  the  uni- 
form rate  of  0.2  inch  per  second.  Find  the  rate  at  which  the  area 
is  increasing  when  the  side  is  10  inches  long. 

Let  X  be  the  length  of  the  side,  and  y  the  area  of  the  square. 

dx  dv 

Then  ^rr  =  0.2  and  ^r  is  the  rate  of  increase  of  the  area.     To  find 
dt  dt 

this  rate  of  increase,  differentiate  the  function  y  =  x^. 

^  -2x— • 
dt       ^^  dt 

Since 

dx 

dt   =  0-2. 

dt       "-^ 


§31]  THE  POWER  FUNCTION  39 

dv 
When  X  =  10,  -rr  =  4.     The  area  is  increasing  at  the  rate  of  4 

dv 
square  inches  per  second.     When  x  =  13,  -tt  =  5.2,  the  rate  of 

change  of  the  area  at  this  instant. 

Illustration  2.     A  spherical  soap  bubble  is  being  inflated  at  the 

rate  of  0.2  cubic  inch  per  second.     Find  the  rate  at  which  the 

radius  is  increasing  when  it  is  1.5  inches  long. 

dV 
Let  r  be  the  radius,  and  V  the  volume  of  the  bubble,     -rr  =  0.2 

at 

dv 
and  ^7»  the  rate  of  increase  of  the  radius,  is  to  be  found. 

V  =  AirrK 


^  =  ^^^^41- 

From  which 

dr        1     dV 

dt  ^  47rr2   dt 

Since 

dV 

TT-  =  0.2,  and  r  =  1.5, 

dr  1 

tit  ~  on-n  K\2  ~  0.0071  inch  per  second. 

Exercises 

1.  Find  the  rate  at  which  the  surface  of  the  soap  bubble  of  Illustra- 
tion 2  is  increasing  when  r  =  1.5  inches. 

2.  If  each  side  of  an  equilateral  triangle  is  increasing  at  the  rate 
of  0.3  inch  per  minute,  at  what  rate  is  the  area  of  the  triangle  increas- 
ing when  the  side  is  6  inches  long? 

3.  Water  is  flowing  at  a  uniform  rate  of  10  cubic  inches  per  minute 
into  a  right  circular  cone  whose  semi-vertical  angle  is  45°,  whose  apex 
is  down,  and  whose  axis  is  vertical.  At  what  rate  is  the  surface  of 
the  water  in  the  cone  rising,  and  at  what  rate  is  the  area  of  this  surface 
increasing  when  the  water  in  the  cone  is  25  inches  deep? 

31.  The  Derivative  of  the  Sum  of  a  Function  and  a  Constant. 

Sketch,  on  the  same  set  of  axes,  the  graphs  of  the  functions: 


40  CALCULUS  [§31 


y  =  x^;  y  =  x^  —  5;  y  =  x^  -{-  3;  y  =  x^  -\-  10.     Find  -v-  for  each 

dy 
of  the  functions.     Find    t-,   if    y  =  x"^  -{-  C,    where    C    is    any 

constant. 

Sketch  a  graph  of  any  function  y  =  f{x),  and  on  the  same  set 
of  axes,  graphs  oi  y  =  fix)  -\-  C  for  several  values  of  the  constant 

dy 
C.     What  relation  exists  between  t-  for  the  different  functions 

dx 

corresponding  to  the  same  value  of  cc? 

From  these  illustrations  it  is  clear  that  the  derivatives  of  all 
Junctions  which  differ  only  by  an  additive  constant  are  the  same. 
The  reason  for  this  is  geometrically  evident.  For,  the  addition 
of  a  constant  to  a  function  has  the  effect  of  merely  translating  the 
graph  of  the  function  parallel  to  the  F-axis.  The  slope  corre- 
sponding to  any  given  abscissa  is  clearly  not  changed  by  this 
translation.     Hence, 

d[f(x)+C]       d[f^x)l 


dx  dx 

In  particular        d[axn-|-C]       d[ax'i] 

_ =  _- —  =  a 

dx  dx 

Thus,  if  _  ,   .    „ 

dy       d{5x^) 


rll— 1 


(1) 

(2) 


dx  dx 

Exercises 


15x2 


1.  Prove  formula  (2)  above  by  the  increment  process. 
Differentiate : 

2.  y  =  3x^  +  2.  11.  7/  =  -  4x^  +  6. 

3.  y  =  5\/x  +  4.  12.  y  =  -  3x*  +  2. 

4.  y  =  2x3  _  3.  13.  y  =  Tx^  -  3. 

3 

5.  2/  =  — 5  +  5.  14.  ?/  =  4x'  +  5. 

6.  2/  =  2t*  +  7.  15.  2/  =  ^  +  2. 

7.  s  =  16<2  +  5.  16.  2/  =  -  ix«  +  3. 

8.  «  =  2y/t^  +  6.  17.  2/  =  ix*  +  2. 

9.  s  =  ^3  -  4.  18.  2/  =  ?x*  -  5. 
10.  X  =  4<3  -  2. 


§32]  THE  POWER  FUNCTION  41 

32.  The  Derivative  of  au°.     li  y  =  aw,  where  w  is  a  function 
of  X  and  n  is  a  positive  integer,  the  student  will  prove,  as  in  §30, 

that 

dy  .du 

dx  dx 

or 

d(au'')  „  .du 

dx  dx 

Illustrations. 

=  30x(x2  +  3)2. 

2    d[2(x''  +  4)^+10]_rf[2(x2  +  4)^]_,^  ,(J(x2  +  4) 

dx  dx  dx 

=  2-4(a;2  +  4)3  2.r  =  16x(x2  +  4)'. 

3.  If  y  =  (2x2  +  1)2,  find  ^. 

dy  _  cj(2x2+l)2  _  ,    1^  ^(2x2+  1) 


•2'2x 
4.  If  77  =  (x2  +  l)t. 


=  2(2x2  +  i).2-2x  ^  =  8x(2x2  +  1)  ^j- 


dy  _  3 
dx  ~  " 

(a:2+l)^ 

2x  = 

3x(x2  +  1)5, 

and 

dy      3 

(X2+1)^ 

dt 

=  3x(x2 

+  1)^ 

dx 
dt' 

Exercises 

Fi 

nd^: 
ax 

y  =  (4x2 

y  =  5(2x 

y  =  2(3 
y  =  \/x 
y  =  (5- 

-2)3. 
2    -   5)3. 

-  4X2)3 

-  X2)3. 

6.  y 

7.  y 

8.  y 

9.  y 

10.  y 

1. 
2. 
3. 
4. 
6. 

=  V9 
=  (3x2 
=  a/x^ 
=  V2a 

-X2. 

+  7)2. 
-5. 

;2  +  3 
;2  +  l. 

42  CALCULUS  [§33 


11-  y  =      /^         '•  15.  y  = 


VSx  -  2  A^x2  +  1 

7  "i 

12.  2/  =  -7===-.  16.  y  = 


2 


13.  y  =  V(x«  +  4).  17.  y  =  y- 


14.  2/  =  V(x  +  1).  18.  2/  = 


(a:2  +  4)2 
3 


(5  -x»)2 

33.  The  Derivative  of  u*^,  n  a  Positive  Fraction.     We  are  now 

in  a  position  to  prove  that  the  rule  for  the  derivative  of  m"  holds 

when  n  is  a  positive  fraction  of  the  form  -'  where  p  and  q  are 

integers.     Let 

p 
y  =  w. 

Raise  each  member  to  the  power  q: 

?/«   =   UP. 

Since  m  is  a  function  of  x,yisa  function  of  x.  Hence  each  member 
is  a  function  of  x  raised  to  a  positive  integral  power.  Then  each 
member  can  be  differentiated  by  the  rule  of  §32  which  was 
proved  for  positive  integral  exponents.     We  find 


,  dy 

^y-'  Tx  = 

=  pu 

-1  '^'^. 
dx 

From  which 

dy      V 

tip-i 

'du 

dx       q  y-^ 

dx 

Substitute  for  y  in 

the  second  member  and  obtain 

dy 

P 

wp-^ 

du 

p  up- 

-1  du 

dx  ~ 

9 

u' 

dx 

q 

up- 

p  dx 

Then 

dy  ^1 
dx       q 

u 

du 
dx 

§35]  THE  POWER  FUNCTION  43 

and  the  rule  is  proved  that 

dw  ,  du 

ax  dx 

where  n  is  a  positive  fraction  whose  numerator  and  denominator 
are  integers.  This  rule  has  already  been  used  in  the  solution  of 
numerous  exercises. 

34.  The  Derivative  of  a  Constant.    Let  y  =  c,  where  c  is  a 
constant.     Corresponding  to  any  Ax,  Ay  =  0,  and  consequently 

Ax       "' 


lim    -^  =  0. 


and 

Ax  =  0   Ax 

or 

^  =  0. 
dx 

The  derivative  of  a  constant  is  zero. 
Interpret  this  result  geometrically. 
35.  The  Derivative  of  the  Sum  of  Two  Functions.    Let 

y  =  u  +  v, 

where  u  and  v  are  functions  of  x.  Let  Au,  Av,  and  Ay  be  the  incre- 
ments of  u,  V,  and  y,  respectively,  corresponding  to  the  increment 
Ax. 

y  -\-  Ay  =  u  -{-  Au  +  V  -jr  Av 

Ay  =  Au  -\-  Av 

Ay  _  Au      Av 

Ax  ~  Ax      Ax 

dy       du    .   dv 

dx       dx       dx 
or 

d(u4-v)  _  du      dv 

dx       ~  dx       dx* 

The  derivative  of  the  sum  of  two  functions  is  equoiL  to  the  sum  of 
their  derivatives. 


44  CALCULUS  [§36 

The  student  will  observe  that  the  proof  given  can  be  extended 
to  the  sum  of  three,  four,  or  any  finite  number  of  functions. 
Illustrations. 

I   t^(63^  +  15a;')  ^  d(6x)       rf(15x')  _  g   ,   30^ 
dz  dx  dx 

^   d{2Vx  +  3x'  +  4)      d{2V~x)    ,   d(3x^)    ,  d(4)         1 

2.  = h. ' =  — 7=  +  ox. 

rfx  dz  dx  dx       y/x 

d(t^  +  2<'  +  3) 

3.  f;  =  2<  +  6«2 

Exercises 

Differentiate  the  following  functions  with  respect  to  x,  also  with 
respect  to  t: 

1.  3x*  -  2x^  +  6.  10.  ax2  +  bx  +  c. 

2.  5x»  -  7x2  -2x  -  10.  11.  y  =  v^a;2  +  4^;  _  5. 

3.  gx*  -  ix'  +  X  -  7.  12.  2/  =  1 


4.  x2  +  2x 
3  2 


y/x^  -  5x  +  7 


13.  2/  =  V3x2  _  2x  +  5. 

*  \7x  ~  ^"  ^         14.  y  =  Ve  -  3x  -  x\ 

6.  3x7  _  63.8  +  9.  16.  s  =  Vm  +  ^2<  -  3 

?_2                                                                                              1 
..   -.    —  X    ^  16.    7/   = 

x"  -  7x  -  6 

8.  -  \x^  -\-  \x^  -  X  +  2.  _  / 

II  17.  y  =  V  a;^  -  5x  +  4. 

9.  x~5  4-  a;~5.  18_  y  _  (3_j.2  _|.  2x  +  2)3. 

36.  Differentiation  of  Implicit  Functions.  The  derivative  of 
one  variable  with  respect  to  another  can  be  found  from  an  equa- 
tion connecting  the  variables  without  solving  the  equation  for 
either  variable.  For,  if  the  variables  are  x  and  t/,  j/  is  a  function 
of  x,  even  though  its  explicit  form  may  not  be  known,  and  the 
usual  rules  for  finding  the  derivative  of  functions  can  be  applied 
to  each  member  of  the  equation. 

The  following  example  will  illustrate  the  process. 

dy 
Illustration.     Letx'  +  2/^  =  O"-     Find  -?  • 

The  left-hand  member  of  the  given  equation  is  the  sum  of 


§36] 


THE  POWER  FUNCTION 


45 


two  functions  of  x,  since  y  is  a  function  of  x.  Further,  the  deriva- 
tive of  the  left-hand  member  is  equal  to  the  derivative  of  the 
right-hand  member.  The  derivative  of  the  latter  is  in  this  case 
zero,  since  the  right-hand  member  is  constant.     On  differentiat- 


x+  y=^o, 


Fig.  24. 


ing  the  left-hand  member  as  the  sum  of  two  functions,  we  obtain 


Solving  for  -t-> 


2x  +  2y%  =  0. 


dx 


When  the  derivative  is  found  by  differentiating  each  member  of 
an  equation  in  the  implicit  form,  as  in  the  foregoing  illustration, 
the  operation  is  called  implicit  differentiation. 


46  CALCULUS  [537 

Exercises 

1.  Draw  the  circle  x^  +  y''  =  o*  and  show  geometrically  that  the 
slope  of  the  tangent  at  the  point  (x,  y)  is 

dy 

2.  Solve  the  equation  of  Exercise  1  for  y  and  find  -j-  • 

dv 
From  the  following  equations  find  -r  by  implicit  differentiation: 

3.  3x«  +  4y*  =  12. 

4.  a;^  —  ?/*  =  a^. 

^'  ^  +  ^  =  !•     (^o  ^'^^  clear  of  fractions.) 

If  y  is  an  implicit  function  of  x  expressed  by  an  equation  of  the 
form 

x"  +  y  =  a",  (1) 

differentiation  gives 

or 

g=-[i]""'  <^> 

The  equation  (1)  includes  a  number  of  important  special  cases. 
The  graphs  corresponding  to  the  following  values  of  n  are  shown  in 
Fig.  24.     For 

n  =  h    x'  +  y*  =  o^,  a  parabola, 

2  1  2  .  ■ 

n  —  I,     x^  -\-  y^  =  a^,  an  important  hypocycloid, 

n  =  l,     X     +  y     =a,     a  straight  line, 

n  =  2,    x"^  -\-  y"^    =  o^,  a  circle. 

The  graph  of  (1)  passes  through  the  points  (0,  a)  and  (a,  0)  if  n  is 
positive. 

12? 

6.  x»  +  J/*  =  o^. 

Ill 

7.  x*  +  2/2   =  o2. 

8.  x'  +  y'   =  a\ 

3  3  3 

9.  x^  +  y^  =  a^. 

37.  Anti-derivatives.  Integration.  Let  it  be  required  to 
find  the  equation  of  a  curve  whose  slope  at  any  point  is  equal  to 
twice  the  abscissa  of  that  point. 


§37]  THE  POWER  FUNCTION  47 

dy 
This  means  that  at  every  point  of  the  curve  -,     =  2x.     We 

seek  then  a  function  whose  derivative  is  2x.  y  =  ^^  is  such  a 
function.  But  y  =  x^  -\-  C,  where  C  is  a  constant,  is  also  a 
function  having  the  same  derivative.  Hence  there  is  an  infinite 
number  of  functions  whose  derivatives  are  all  equal  to  2x.  The 
problem  as  proposed  has  then  an  infinite  number  of  solutions, 
viz.,  the  system  of  parabolas  y  =  x^  +  C,  corresponding  to  the 
infinitely  many  values  of  C. 

If  now  we  add  to  the  statement  of  the  problem  the  requirement 
that  the  curve  shall  pass  through  a  given  point,  say  (1,  2),  it  is 
geometrically  evident  that  but  one  of  the  curves  y  =  x^  +  C  will 
pass  through  the  point.  In  other  words  there  is  but  one  value  of  C 
for  which  the  latter  requirement  is  satisfied.  This  value  is  de- 
termined by  substituting  the  coordinates  of  the  point  in  the 
equation  y  =  x^  -{-  C,  since  they  must  satisfy  this  equation  for 
some  value  of  C,  if  the  problem  has  a  solution.  On  making  the 
substitution  we  have 

2  =  1  +  C, 

from  which  (7=1.  Hence  y  =  x"^  -\-  1  is  the  equation  of  the 
curve  whose  slope  at  any  point  is  equal  to  twice  the  abscissa  of  the 
point  and  which  passes  through  the  point  (1,  2). 

The  nature  of  the  problem  which  has  just  been  solved  can  be 
further  explained  by  the  following  geometrical  solution.  Draw, 
Fig.  25,  at  the  vertices  of  each  small  square  on  a  sheet  of  co- 
ordinate paper  on  which  a  set  of  axes  has  been  chosen,  short 
lines  whose  slopes  are  equal  to  two  times  the  abscissas  of  the 
respective  vertices.  A  curve  is  to  be  drawn  which  at  each  of  its 
points  is  tangent  to  a  line  such  as  those  which  have  been  drawn. 
Now  it  is  impossible  in  the  figure  to  draw  lines  through  every  point 
in  the  plane,  but  if  the  points  through  which  the  lines  are  drawn 
are  sufficiently  thick,  the  lines  will  serve  to  indicate  the  direction 
which  the  curve  takes  at  nearby  points.  The  lines  may  be 
regarded  as  pointers  indicating  the  stream  lines  in  flowing  water. 
Then  a  point  tracing  the  curve  would  move  as  a  small  cork  would  in 
water  having  the  stream  lines  indicated  by  the  figure. 

Thus,  to  get  the  curve  that  goes  through  (1,  2),  start  from  this 
point  and,  guided  by  the  direction  lines,  sketch  in  as  accurately 


48 


CALCULUS 


I§37 


as  possible  the  curve  to  the  right  of  this  point.  Do  the  same 
thing  to  the  left,  noting  that  here  it  is  necessary  to  go  against  the 
stream  lines  instead  of  with  them. 

In  Fig.  25  it  should  be  noted  that  all  lines  through  points 
having  the  same  abscissa  are  parallel.    This  fact  is  of  great 


^      ^      ^      ^     ^     \ 


^ 


M   ^  ^  H  ^ 


k   k   h  \  \  ^ 
k   k   k   \^\^ 


/MM 


/^  ^  M  M 


Fig.  25. 

assistance  in  drawing.  The  squares  on  the  coordinate  paper  can 
be  used  to  advantage  in  drawing  lines  when  the  slope  is  known. 

If  the  derivative  which  was  given  had  been  any  other  function 
of  X,  a  geometrical  solution  could  have  been  obtained  by  the  same 
method.  - 

The  foregoing  illustration  introduces  a  new  type  of  problem, 


§37]  THE  POWER  FUNCTION  49 

viz.,  that  of  finding  a  function  whose  derivative  is  given.  A 
function  whose  derivative  is  equal  to  a  given  function  is  called 
an  anti-derivative,  or  integral,  of  the  given  function.  From  the 
illustration  it  is  clear  that  any  given  function  which  has  one  anti- 
derivative,  has  an  infinite  number  of  anti-derivatives  which  differ 
from  each  other  only  by  an  additive  constant.  This  latter  fact  is 
indicated  in  obtaining  the  anti-derivative  of  a  given  function  by 
writing  down  the  variable  part  of  the  anti-derivative  and  adding 
to  it  a  c  onstant  C  which  is  undetermined  or ' '  arbitrary. ' '  In  a  given 
application  this  constant  will  be  determined  by  supplementary 
conditions  as  in  the  illustration  at  the  beginning  of  this  section. 
The  process  of  finding  the  anti-derivative  of  a  given  function  is 
called  integration. 


y  =  x^  +  C. 

y  =  lx'  +  C. 

y  —  x^  ■{-  x^  -\-  C. 

y  =  x^  +  x"^  -\-  Ix  -\-  C. 

2/  =  3-  +  2-  +  7a;  +  C. 

If  in  Illustration  1  the  curve  is  to  pass  through  the  point 
(3,  -2)  we  must  have  -2  =  3'  +  C,  or  C  =  -29.  Hence  the 
equation  of  the  curve  is  y  =  x^  —  29. 

Exercises 


Illustrations. 

1. 

jfdy 
dx  ~ 

3x2 

2. 

If  —  - 
dx 

x\ 

3. 

-ffdy 

3x2 

+  2x, 

4. 

If  —  - 
dx 

3x2 

+  2x  -h  7, 

5. 

Tf  ^  - 
dx 

x^  +  x  +  r. 

Integrate  the  following  ten  functions: 

1- 1  =  3,. 

^-  dt  ~ 

dx 

(3x2  +  2x  +  6)  ^ 

..  g  =  4... 

^    dy  _ 
''  dt 

(ax  +  b)  ^-. 

''•|--4r 

^-  dx~ 

3x2  _  2x2  _|.  7. 

«•  f  =  3..f . 

Q  '^y 

^'  dx- 

lOx-2  +  2x-3  -  X 

6.  ^  =  3x2  +  2x  -  6. 

10.  f~  = 

dx 

x^  +  x'\ 

60  CALCULUS  l§37 

Illustrations. 

6.       ^  =  3(x2  +  2)2  2x. 

The  right-hand  side  is  in  the  form,  nw-^  ^,  where  n  is  3,  and  w 

du  -, 

is  (x'^  +  2).     Since  the  integral  of  nw"-i  ^  is  -u"  +  C, 

2/  =  (x^  +  2)»  +  C. 

7.  ^  =  (x2-5)'2x  =  H4(x2-5)'2x]. 

2/  =  i(x2  -  5)^  +  C. 

8.  ^  =  x(x2-  l)''  =  -iM6(x2-  l)''2x]. 

y  =  Mx^  -  1)«  +  C. 

y  =  _  -j^(3  -  x')«  +  (7. 
10.     I  =  (.'  -  2x  +  3)-'(x  -  l);j^ 

=  -  il  -  2(x^  -  2x  +  3)-^(2x  -  2)^J. 

y  "  4(x2  -  2x  +  3)2  +  ^* 

Exercises 
Integrate : 

11.  ^  =  a;V^^"^=l-     Ans.  y  =  K^^  -  1)^  +  C. 

12.  ^  =  (2x3  4.  3x2) 3 (x^  _|_  x).     Ans.  y  =  i(2x3  4.  3x2)1  +  c, 
ax 


13.  ^  =  (x  +  1)*.     ^ns-  y  =  f  (x  +  1)^  +  C. 


dx 


14.  I  =  (2- XT  X.  16.  ^^  MX- 3)2  X. 

16.|  =  xVr^'.  17.g  =  x2+3x. 

18.^^  =  (x2  +  7)3x. 


§37]  THE  POWER  FUNCTION  51 

20.  f^  =  (x^  +  2x  +  iy(x  +  1). 


22. 

dy 

dx  ~ 

X 

24 

dy 
dx 

X2 

V5 

-X* 

-  (4  -  x")* 

25. 

dy 

dx  ~ 

(3x2 

+  2) 

'  X.     Ans. 

y 

=  A 

(3x2 

+  2y  +  c. 

26. 

dy 

dt  ~ 

(2- 

-  3x2) 

.    dx 
'"'dt- 

27. 

dy  ^ 
dx 

(2x 

+  1)^ 

Ans.  y 

= 

K2x 

+  1] 

3  +  C. 

28. 

dy 
dt   ~ 

(3x 

-2y 

dx 
dt' 

29.^f  =  (3-4x)2^. 

30.  ^  =  V^TT-     ^ns.  2/  =  f (x  +  1)'  +  C. 

31.  ^  =  >/2F+^.  40.  ^  =  \/(2x+3)». 

33.  -2  =  Vir=^.  42.  ^  =  Vx^  +  3x  -  7  (x^  +  1). 

3*-  5x  =  vTTT'  "•  ^'  ^  ^"  "^  4)^/iM^8^+^. 

^^-  ^^  =  virr2  '*•  ^'  =  ^^  -  ^)^^^^^+^- 

^^    dy              1  .-    dy                1  —  5x 

36.  T^  =      ,  46.     " 


dx       V3  -  5x  *  dx       -y/e  4-  4a;  -  lOx* 

—    dw  / .^    dy  3x  —  2 

37.  /  =  X  V4x2  _  5.  46.     *' 


dx  *  '  dx       V3x2  _  4a;  _|_  5 

88.  I  -  .V9^'.  «.  I  -  V4^. 

89.  f  .  '^ 48.  f^  -  x(2  -  x>).. 

dx       .^^4  —  3x2  dx 

49.  Find  the  equation  of  the  curve  whose  slope  at  any  point  is  equal 
to  the  square  of  the  abscissa  of  that  point  and  which  passes  through 
the  point  (2,  3). 

60.  Find  the  equation  of  the  curve  whose  slope  at  any  point  is  equal 


52  CALCULUS  I  §38 

to  the  square  root  of  the  abscissa  of  that  point  and  which  passes 
through  the  point  (2,  4). 

51.  Find  the  equation  of  the  curve  whose  slope  at  any  point  is  equal 
to  the  negative  reciprocal  of  the  square  of  the  abscissa  of  that  point 
and  which  passes  through  the  point  (1,  1). 

38.  Acceleration.  The  velocity  of  a  body  moving  in  a  straight 
line  may  be  either  uniform  or  it  may  vary  from  instant  to  instant. 
In  the  latter  case  its  motion  is  said  to  be  accelerated,  and  this 
applies  both  to  the  case  where  there  is  an  increase  in  velocity 
and  the  case  where  there  is  a  decrease  in  velocity. 

Thus  it  is  a  fact  of  common  knowledge  that  the  velocity  of  a 
body  falling  to  the  ground  from  a  height  increases  with  the  dis- 
tance through  which  the  body  has  fallen,  or  with  the  time  since 
the  body  started  to  fall.  The  time  rate  of  change  of  the  velocity 
of  a  moving  body  is  an  important  concept  in  mechanics  and 
physics. 

If  s  denotes  the  distance  passed  over  in  time  t,  the  velocity  has 
been  defined  as  the  rate  of  change  of  s  with  respect  to  t.  The 
notion  of  the  velocity  at  a  given  instant  was  derived  from  that  of 
the  average  velocity  for  an  interval  A^  The  average  velocity  was 
obtained  by  dividing  the  change  in  s,  As,  in  a  time  A^  by  At  {i.e., 
by  dividing  the  distance  passed  over  in  time  At  by  At).  The 
limiting  value  of  this  quotient  as  At  approaches  zero  was  defined 
as  the  velocity  at  the  beginning  of  the  interval  A^. 

In  the  same  way  if  the  velocity,  r,  changes  by  an  amount  Av  in 
the  time  A^  counted  from  a  certain  time  t,  the  average  rate  of 

Av 
change  of  v  for  this  interval  is  ^-     It  is  the  average  linear  accelerch 

tion^  for  this  interval.  The  acceleration  at  the  time  t  is  defined  as 
the  limit  of  the  average  acceleration  as  At  approaches  zero.     It  is 

then  ^'     The  acceleration  is  the  time  rate  of  change  of  velocity.     In 

the  case  of  a  falling  body  it  is  known  experimentally  that  for 
bodies  falUng  from  heights  that  are  not  too  great,  the  velocity 
changes  uniformly,  due  to  the  action  of  the  force  of  gravity,  i.e., 

•  We  suppose  here  that  the  body  is  moving  in  a  straight  line.  If  the  path  is  curved, 
it  will  be  seen  later  that  the  total  acceleration  is  to  be  thought  of  as  the  resultant  of 
two  components,  one  of  which  produces  a  change  in  the  direction  of  the  velocity  and 
the  other  a  change  in  the  magnitude  of  the  velocity. 


§38] 


THE  POWER  FUNCTION 


63 


the  time  rate  of  change  of  the  velocity  is  a  constant.  This  con- 
stant is  called  the  acceleration  due  to  gravity  and  is  usually 
denoted  by  g.  In  F.P.S.  (foot-pound-second)  units  it  is  equal 
to  32.2  feet  per  second  per  second.  That  the  unit  of  acceleration 
is  1  foot  per  second  per  second  is  explained  by  the  fact  that  accelera- 


tion is  the  change  per  Second  of  a  velocity  of  a  certain  number  of 
feet  per  second. 

The  differential  equation  of  motion  of  the  falling  body  can  be 
written 

=  9-  (1) 


dt 


From  which  on  integrating, 


V  =  gt  +  C. 


If  it  is  given  that  the  body 
starts  falling  from  rest,  we  have 
as  the  condition  for  determin- 
ing C,  that  V  =  0  when  t  =  0. 
Equation  (2)  shows  that  C  must 
be  equal  to  zero.     Then, 


V  =  gt. 


(3) 


(2) 


Time  U) 

Fig.  27. 


The  graph,  Fig.  27,  o(  v  =  gt  is  a,  straight  line  whose  slope  is  g. 

If  the  body  had  had  an  initial  speed  of  Vq  feet  per  second,  i.e., 

if  it  had  been  projected  downward  instead  of  being  dropped,  the 

constant  C  would  have  been  determined  from  the  condition  that 


54  CALCTJLUS  [§38 

V  =  Vo  when  t  =  0.  It  follows  from  (2)  that  C  =  vo,  and  the 
equation  for  v  would  have  been 

V  =  gt  +  Vq.  (4) 

The  graph  of  this  function  is  shown  in  Fig.  26.  It  is  again  a 
straight  line  but  it  cuts  the  F-axis  at  the  point  (0,  Vo). 

The  foregoing  discussion  evidently  applies  equally  well  to  any 
uniformly  accelerated  motion,  i.e.,  to  any  motion  where  the  rate  of 
change  of  the  velocity  is  constant.  In  all  such  cases  the  graph  of 
r  as  a  function  of  the  time  is  a  straight  line. 

Since  v  =  -n'  equation  (4)  gives 

ds 

di  =  9i  +  f  0. 

Integrating, 

s  =  yt^  +  vot  +  Ci.  (5) 

If  t  is  measured  from  the  instant  the  body  begins  to  move  and  s 
from  the  position  of  the  body  at  that  instant,  s  =  0  when  t  =  0. 
From  this  condition  d  =  0.  Then  the  distance  of  the  body  from 
its  initial  position  is  given  by 

s  =  igf^  +  Vot. 

If  a  body  is  thrown  vertically  upward,  it  is  convenient  to  count 
distances  measured  upward,  and  upward  velocities,  as  positive. 
Then,  since  the  acceleration  due  to  gravity  diminishes  v,  equation 
(1)  becomes 

di  =  -^-  (^  ^ 
The  formulas  (4)  and  (5)  then  become 

v^  -gt  +  vo  (4') 

s  =  -  hgt'  +  Vot.  (5') 

If  a  body  falls  from  rest  it  is  easy  to  express  the  speed  as  a 

function  of  the  distance  traversed.     In  this  case,  2^0  =  0.     Then 

(4)  and  (5)  become, 

V  =  gt  (4") 


i 


§38]  THE  POWER  FUNCTION  55 

Elimination  of  t  between  these  equations  gives: 

V  =  VWs-  (6) 

Exercises 

1.  If  a  body  falls  from  rest  how  far  will  it  fall  in  10  seconds? 

2.  If  a  body  is  thrown  vertically  downward  with  a  velocity  of  10  feet 
per  second,  how  far  will  it  have  moved  by  the  end  of  10  seconds? 
What  will  its  velocity  be? 

3.  If  a  body  is  thrown  vertically  upward  with  a  velocity  of  64.4 
feet  per  second,  what  will  the  velocity  be  at  the  end  of  10  seconds? 
What  will  be  the  position  of  the  body?     How  far  will  it  have  moved? 

4.  Find  the  laws  of  motion  if  the  acceleration  is  equal  to  2t  and  if 
(1°)  s  =  0  and  v  =  0  when  t  =  0;  (2°)  s  =  3  and  v  =  --2  when  t  =  0. 

6.  If  the  acceleration  is  proportional  to  the  time  and  ii  v  =  vo  and 
8  =  So  when  i  =  0,  show  that 

8    =   -^    +  Vol   +  So. 


CHAPTER  IV 
DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

39.  The  Derivative  of  the  Product  of  a  Constant  and  a  Variable. 

Let 

y  =  cu, 

where  c  is  a  constant  and  u  and  y  are  functions  of  x.  Let  Am  and 
Ay  be  the  increments  of  u  and  y,  respectively,  corresponding  to 
the  increment  Ax.     Then 


2/  +  A?/  = 

c{u  +  Am) 

A?/  = 

cAm 

Ay 

Ax 

Am 
Ax 

dy 

Ix  ~ 

du 

c  v-» 

dx 

d(cu) 
dx    ~ 

du 

C  J* 

dx 

or 


The  derivative  of  the  product  of  a  constant  and  a  function  is  equal 
to  the  constant  times  the  derivative  of  the  function. 
Illustrations. 

d{3x^)  d(x2) 

1.     — J =   O    —j — ■•  =   DX. 

ax  dx 


,m__m,_,^j^._,,^_,. 


3. 


d[-f(x2-5)2] 

dx 


,d{x^ 

^-5)^ 

-i  d{x^  - 
dx 

= 

3 
2 

dx 

2-5)- 

X 

A) 

3 

(X2- 

50 

-5)2 

§40]  ALGEBRAIC   FUNCTIONS  57 

40.  The  Derivative  of  the  Product  of  Two  Functions.     Let 

y  =  uv, 

where  u  and  v  are  functions  of  x. 

y  -\-Ay  =  (u  +  Au)(v  +  Aw) 
y  +  Ay  =  uv  -\-  nAv  +  vAu  +  AuAv 
Ay  =  uAv  +  vAu  +  AuAv 

Ay         4!i_i_  4!f_i_A  — 
Ax  ~  Ax         Ax  Ax 

Since  Aw  approaches  zero  as  Ax  approaches  zero, 

dy  _  dv  du 

dx  dx         dx 
or 

d(uv)  dv  ,       du  ... 

^^dr  =  ^dx  +  ^dx'  ^^^ 

The  derivative  of  the  product  of  two  functions  is  equal  to  the  first 
times  the  derivative  of  the  second  plus  the  second  times  the  derivative  of 
the  first. 

Illustraiions. 

-  ^J^i^  =  (.  +  „^J^  +  (.  +  3,^i^t^ 
=  (i  +  2)  +  (I  +  3)  =  2i  +  5. 

=  (x2  +  3x)  +  (x  -  2)(2x  +  3) 
=  3x2  _|_  2x  -  6. 
3.  If  x^  +  xy^  +  y  =   10, 

dx      ^y   ^dx 
2x  +  3xy^f^  +  y^  +  f^  =  0, 

whence 

dy  ^  _  2x  +  y»  ^ 
dx     3xy»  +  l' 


58  CALCULUS  [§41 

Exercises 
Differentiate  the  following: 


1.  (x  +  l)ix  -  1).  9.  2/  =  (x  -h  l)VJi^^. 

2.  (x^  +  2x)(x  -  3).  10.  y  =  (2  -  x)\/^c^~:^. 

3.  a;(x2  +  2x  -  6).  11.  y  =  {2x  ■{■  3)\/4  -  x\ 

4.  (x  -  l)='(x^  +  1).  12.  ?/  =  x\/n^. 

5.  (x'  +  2x  -  3)(x  +  1)2.  13.  xV  +  3x  -  7?/  =  15 

6.  xVx  -  1.  14.  a;2?/  -  Sxy^  =  lo. 

7.  (x  -  1)V^.  15.  yy/^  +  xv^  =  3. 

8.  x(x  -  l)i  16.  xy  -  xhj  =  0. 

41.  The  Derivative  of  the  Quotient  of  Two  Functions.    Let 

u 

y  =  -y 

where  u  and  v  are  functions  of  x.     Then 

yv  =  u. 
Differentiating  by  the  rule  of  §40, 

dv  dy  _  du 

^  dx         dx  ~  dz 

du  dv 
dy  _  dx  ^dx 
dx  V 


Replacing  y  by  its  value,  -> 


(v) 


du         dv 
Vx  ~  ^dx  (1) 


dx 


The  derivative  of  the  quotient  of  two  functions  is  equal  to  the 
denominator  times  the  derivative  of  the  numerator  minus  the  numera- 
tor times  the  derivative  of  the  denominator,  all  divided  by  the  square 
of  the  denominator. 


§42]  ALGEBRAIC  FUNCTIONS  59 

Illustration. 

47^72-]  _  (^  -  2)  -^^ (x^  +  1)  -^^- 

dx  (a; -2)2 

_  ix-2)i2x)  -  (x^+1) 
(a: -2)2 
x^  —  Ax  —  1 


{x  -  2)2 

Exercises 

Differentiate  the  following: 
1.  ^-  4.  ^^-r-'.  7. 


a;-l  1+a-  x  -  Vx^  -  1 


V'x  -  1 

1  -\-x 

x^  -  1 

Vx 

x 

x2  -3  a:2  -  1  g2  4,4 

X  —  2  V  X  •''  —  2 

„    X  +  x'^                        „           a;                         „a;3  +  8 
3.  ~~ 6.  T--  9.  hi' 

1  -  a;  (1  -  x)^  ^  ~  ^ 

42.  The  Derivative  of  u",  n  Negative.     In  Chapter  III  the 
formula 

dw"  ,  du 

dx  dx 

was  proved  for  n  positive  and  commensurable.  The  formula 
was  assumed  for  negative  exponents.  We  are  now  in  a  position 
to  give  a  proof  of  the  formula  for  this  case.     Let 


y  =  u-% 

where  s 

is  a  positive  commensurable  constant. 

Then 

1 

or 

yw  =  1. 

Differentiate  by  the  formula  of  §40, 

du            dy 
ysu'-^^^+u-^^  = 

0, 

dy            sy  du 

= ■    J-  =  —  su- 

dx            u   dx 

8-1    ^'^ 

dx 

(1) 


60  CALCULUS  [§43 

This  completes  the  proof  that 

du°  ,  du 

dx  dx 

if  n  is  a  commensurable  constant.^ 
We  see  from  equation  (1)  that 


'(J) 


nc  du 


dx  u°+i  dx 

Illustration. 

d         S  6         dix^  -  1)  12a; 


(2) 


dx  (x2  -  1)2  (a;2  -  1)»        dx  {x^  -  lY 


2.       * 


Exercises 

llo 
4. 

wmg: 

3 

X  -  1 

7         6     . 

'•  x^  +  1 

6. 

5 

(X  +  ly 

(x^  +  D* 

6 

3 

0.          '      . 

Vx  -  1 

(1-x^y 

1  -X 

Vx 

43.  Maximum  and  Minimum  Values  of  a  Fvmction.  In  Chap- 
ter I  it  was  shown  that  the  derivative  of  a  function  with  respect  to 
its  argument  is  equal  to  the  slope  of  the  tangent  drawn  to  the 
curve  representing  the  function.  The  derivative  is  positive  where 
the  function  is  increasing  and  negative  where  the  function  ia 
decreasing.  These  facts  enable  us  to  determine  the  maximum 
and  minimum  values  of  a  function. 

Additional  exercises  in  finding  maximum  and  minimum  values 
of  a  function  will  be  given  in  this  section. 

Illustration.     Let 

J/  =  2x3  +  33-2  _  12a;  -  10. 
•  ^  =  6x2  +  Gx  -  12  =  0(3,  _j_  2)  (i  _  1). 

>  It  can  be  shown  that  the  formula  also  holds  for  incommensurable  exponents. 


1843] 
If  xi 


ALGEBRAIC  FUNCTIONS 


61 


If  X  is  less  than  —2,  both  factors  of  the  derivative  are  negative. 
Then  for  all  values  of  x  less  than  —2,  the  derivative  is  positive 
and  the  function  is  increasing.  If  x  is  greater  than  —2  and  less 
than  1,  the  first  factor  of  the  derivative  is  positive  and  the  second 
negative.  Hence,  if  —  2  <  x  <  1,  the  derivative  is  negative  and 
the  function  is  decreasing.  If  x  is  greater  than  1  the  derivative  is 
positive  and  the  function  is  again  increasing. 

The  function  changes  from  an  increasing  to  a  decreasing  func- 
tion when  x  passes  through  the  value  —2,  and  changes  from  a 


decreasing  to  an  increasing  function  when  x  passes  through  the 
value  1.  Hence  the  function  has  a  maximum  value  when  x  equals 
—2,  and  a  minimum  value  when  x  equals  1.  These  values,  10  and 
— 17,  respectively,  are  obtained  by  substituting  —  2  and  1  for 
X  in  the  function.  (See  Fig.  28.)  The  more  important  results 
of  the  above  discussion  are  put  in  tabular  form  below. 


X 

X  +  2 

X  -  1 

dy 
dx 

Function 

X  <   -2 





+ 

Increasing. 

-2  <  I  <  1 

+ 

- 

- 

Decreasing. 

1    <    X 

+ 

+ 

+ 

Increasing. 

X  =   -2 

0 

— 

0 

Max.  value  =  10. 

X  =  1 

+ 

0 

0 

Min.  value  =  —17. 

62  CALCULUS  [§43 

It  is  to  be  observed  that  —  2  and  + 1  are  the  only  values  of  x 
at  which  the  derivative  can  change  sign  and  that  these  are  the 
values  that  need  to  be  examined  in  finding  the  maximum  and 
minimum  values  of  the  function. 

Exercises 

Find  where  each  of  the  following  functions  is  increasing;  decreas- 
ing. Find  the  maximum  and  minimum  values  if  there  are  any. 
Sketch  the  curve  representing  each  function. 

1.  y  ^  xK  6.  2/  =  (x  +  2)(x  -3). 

2.  y  =  x\  1.  y  =  2a;'  -  9x^  +  12x  -  10 

3.  t/  =  -2x\  8.  y  =  x'  -  3x  +  7. 

4.  y  =  x'  +  3x  -  2.  9.  2/  =  x'  +  x2  -  X  -  1, 

5.  2/  =  3x'  -  2x«  —  6.  10.  y  =  —^ 

11.  A  sheet  of  tin  24  inches  square  has  equal  squares  cut  from 
each  corner.  The  rectangular  projections  are  then  turned  up  to  form 
a  tray  with  square  base  and  rectangular  vertical  sides.  Find  the  side 
of  the  square  that  must  be  cut  out  from  each  comer  in  order  that  the 
tray  may  have  the  greatest  possible  volume. 

Hint.  Show  that  the  function  representing  the  volume  of  this  tray 
is  4x(12  —  x)'',  where  x  is  the  side  of  the  square  cut  out. 

12.  In  a  triangle  whose  sides  are  10,  6,  and  8  feet  is  inscribed  a  rec- 
tangle the  base  of  which  lies  in  the  longest  side  of  the  triangle.  Ex- 
press the  area  of  the  rectangle  as  a  function  of  its  altitude.  Find 
the  dimensions  of  the  rectangle  of  maximum  area. 

13.  A  ship  A  is  50  miles  directly  north  of  another  ship  £  at  a 
certain  instant.  Ship  B  sails  due  east  at  the  rate  of  5  miles  per  hour, 
and  ship  A  sails  due  south  at  the  rate  of  10  miles  per  hour.  Show 
that  the  distance  between  the  ships  is  expressed  by  the    function 

V  125/*  —  lOOOi  +  2500,  where  t  denotes  the  number  of  hours  since  the 
ships  were  in  the  position  stated  in  the  first  sentence.  At  what 
time  are  the  ships  nearest  together?  At  what  rate  are  they  sepa- 
rating or  approaching  when  f  =  3?     When  <  =  5?     When  <  =  6? 

14.  The  stiffness  of  a  rectangular  beam  varies  as  its  breadth  and  as 
the  cube  of  its  depth.  Find  the  dimensions  of  the  stiffest  beam  which 
can  be  cut  from  a  circular  log  12  inches  in  diameter. 


§431  ALGEBRAIC  FUNCTIONS  63 

Miscellaneous  Exercises 

Differentiate  tlie  following  twenty-five  functions  with  respect  to  x: 


l.i. 

9.  (3  -  x)^ 

18. 

X2(l    -   X). 

X 

10.  (2  -  x^y. 

19. 

x(i  -  x^y. 

11.  (3  -  x=')-2. 

20. 

(1    -  X3)2(X 

-2)2. 

4.  (X  -  sy. 

12.  (2  -  x*)-". 

13.  (x  +  l)(x  -  2). 

14.  Vi. 

21. 
22. 

X  -  1. 
X2  +  1 
1   -X 

1   +X2 

5.  (x  -  2)^ 

15.  x\ 

23. 

(x  -  l)i 

6.    (x2  -  1)2. 

x-6) 

16.      ]. 

24. 

X 

7.  (x'  -  2x2  + 

Va^  -  x2 

8.  (x  -  2) -3. 

17.  x(x  -  1)2. 

26. 

XVI   -  X2. 

Integrate  the 

following  twenty  expressions : 

26.  ^  =  x3. 

dx 

37.  '^  .-? 

ax      X 

i' 

27.  J  =  x3+x2.  .  38.  J  =  ^^^:^ 

28.  -^-^  =  0.3  +  x2  +  X  +  1.  39.  ^^ ^ 


^^  -  .c    -r  u.    -r  -,  -r  X.  ^^         ^^  _  ^^^^ 

^^    dy         1  .^    dy  \  —  X 

29.  -7^  =  x2.  40.      -  - 


dx  ~  dx       (2x  -  x2)2 

30.  ^=-4-  41.^^-    ^ 


dx       ^x  '  ^^        Vx  +  1 

31.  :7:  ==  a;^  +  x^.  42.  ■;t-  = 


dx  '  '  dx        "v/x^-T 

a..  I  =  (.  -  1,^.  43.  g  -  ^^. 

44.  —-  =  x"i 
dx 

46.  ^  =  X  -  1. 

dx 

46.  Find  ~\i  x^  -  y"^  =  a\ 

^-Ftodgitfa  +  l'-i. 

48.  A  ladder  20  feet  long  leans  against  the  vertical  wall  of  a 


33. 

dy 

dx  ~ 

x(l  - 

X2)2. 

34. 

dy 

dx  ~ 

X2(x3 

-  2)5 

36. 

dy 
dx  ~ 

1 

(1- 

xy 

36. 

dy 

dx  ~ 

1 

X2' 

64 


CALCULUS 


:§43 


building.  If  the  lower  end  of  the  ladder  is  drawn  out  along  the 
horizontal  ground  at  the  rate  of  2  feet  per  second,  at  what  rate  is 
its  upper  end  moving  down  when  the  lower  end  is  10  feet  from 
the  wall? 

Hint.    Let  AC,  Fig.  29,  be  the  wall  and  let  CB  be  the  ladder. 
I-et  AB  =x  and  AC  =  y.     Then 


and 


V 

dy 
dt 


=  \/400  - 


But,  since 


dx 
dt 

dy 
dt 


-  X         dx 
\/400  -  x^  dt' 


■-2X 

^400  -  x^ 


The  negative  sign  of  the  deriva- 
tive indicates  that  the  upper  end  pjQ    29. 
of  the  ladder  is  moving  down. 

49.  Answer  the  question  of  Exercise  48,  ifx=0;x=2;a;  =  15; 
a;  =  20. 

50.  With  the  statement  of  Exercise  48,  find  the  rate  at  which  the 
area  of  the  triangle  ABC,  Fig.  29,  is  increasing  when  the  lower  end 
of  the  ladder  is  5  feet  from  the  wall. 

51.  With  the  statement  of  Exercise  48,  find  the  position  of  the  ladder 
when  the  area  of  the  triangle  ABC,  Fig.  29,  is  a  maximum. 


Fig.  30. 

52.  A  ball  is  dropped  from  a  balloon  at  a  height  of  1000  feet.  Ex- 
press the  velocity  of  its  shadow  along  the  horizontal  ground  as  a 
function  of  the  time,  if  the  altitude  of  the  sun  is  20°. 

Hint.  Let  x.  Fig.  30,  be  the  distance  of  the  falling  body  above 
the  earth.  Let  y  be  the  distance  of  the  shadow  from  a  point  on  the 
earth  directly  under  the  falling  body. 


§43]  ALGEBRAIC  FUNCTIONS  65 

53.  With  the  statement  of  Exercise  52,  find  the  velocity  of  the 
shadow  when  the  ball  leaves  the  balloon;  when  it  is  half  way  to  the 
earth;  when  it  reaches  the  earth. 

64.  A  man  standing  on  a  dock  is  drawing  in  a  rope  attached  to  a  boat 
at  the  rate  of  12  feet  per  minute.  If  the  point  of  attachment  of  the 
rope  is  15  feet  below  the  man's  hands,  how  fast  is  the  boat  moving 
when  13  feet  from  the  dock? 

55.  The  paths  of  two  ships  A  and  B,  sailing  due  north  and  east, 
respectively,  cross  at  the  point  C.  A  is  sailing  at  the  rate  of  8  miles 
per  hour,  and  B  at  the  rate  of  12  miles  per  hour.  If  A  passed  through 
C  2  hours  before  B,  at  what  rate  are  the  two  ships  approaching  or 
separating  1  hour  after  B  passed  through  C?  3  hours  after  B  passed 
through  C?     When  are  the  two  ships  nearest  together? 

56.  Two  bodies  are  moving,  one  on  the  axis  of  x,  the  other  on  the 
axis  of  y,  and  their  distances  from  the  origin  are  given  by 

X  =  31^  -  31  +  1, 
y  =  6t  -  12, 

the  units  of  distance  and  time  being  feet  and  minutes,  respectively- 
At  what  rate  are  the  bodies  approaching  or  separating  when  1  =  2? 
When  t  =  5?     When  are  they  nearest  together? 

57.  A  ship  is  anchored  in  35  feet  of  water  and  the  anchor  cable 
passes  over  a  sheave  in  the  bow  15  feet  above  the  water.  The  cable 
is  hauled  in  at  the  rate  of  30  feet  a  minute.  How  fast  is  the  ship 
moving  when  there  are  80  feet  of  cable  out? 

58.  A  gas  in  a  cylindrical  vessel  is  being  compressed  by  means  of  a 
piston  in  accordance  with  Boyle's  law,  pv  =  C.  If  the  piston  is 
moving  at  a  certain  instant  so  that  the  volume  is  decreasing  at  the 
rate  of  1  cubic  foot  per  second,  at  what  rate  is  the  pressure  changing  if 
at  this  instant  the  pressure  is  5000  pounds  per  square  foot  and  the 
volume  is  10  cubic  feet? 

59.  Water  is  flowing  from  an  orifice  in  the  side  of  a  cylindrical  tan  k 
whose  cross  section  is  100  square  feet.  The  velocity  of  the  water  in 
the  jet  is  equal  to  \/2gh,  where  h  is  the  height  of  the  surface  of  the 
water  above  the  orifice.  If  the  cross  section  of  the  jet  is  0.01  square 
foot,  how  long  will  it  take  for  the  water  to  fall  from  a  height  of  100 
feet  to  a  height  of  81  feet  above  the  orifice? 

60.  At  a  certain  instant  the  pressure  in  a  vessel  containing  air  is 
3000  pounds  per  square  foot;  the  volume  is  10  cubic  feet,  and  it  is 
increasing  in  accordance  with  the  adiabatic  law,  piP-*  =  c,  at  the  rate 
of  2  cubic  feet  per  second.     At  what  rate  is  the  pressure  changing  ? 

6 


66  CALCULUS  [§44 

61.  Water  flows  from  a  circular  cylindrical  vessel  whosa  radius  is 
2  feet  into  one  in  the  shape  of  an  inverted  circular  cone  whose  vertical 
angle  is  60°.  (a)  If  the  level  of  the  water  in  the  cylinder  is  falling 
uniformly  at  the  rate  of  0.5  foot  a  minute,  at  what  rate  is  the  water 
flowing?  (6)  At  this  rate  of  flow,  at  what  rate  will  the  level  of  the 
water  in  the  cone  be  rising  when  the  depth  is  4  inches?  When  it  is 
20  inches? 

62.  A  toboggan  slide  on  a  hillside  has  a  uniform  inclination  to  the 
horizon  of  30°.  A  man  is  standing  300  feet  from  the  top  of  the  slide 
on  a  line  at  right  angles  to  the  slide.  How  fast  is  the  toboggan  moving 
away  from  the  man  3  seconds  after  leaving  the  top?  10  seconds  after 
leaving  the  top?  (Use  formula  for  speed  of  a  body  sliding  down  an 
inclined  plane.     Neglect  friction.) 

If  the  man  is  approaching  the  top  of  the  slide  at  the  rate  of  10  feet 
a  second,  answer  the  same  questions,  it  being  supposed  that  the  man  is 
300  feet  away  from  the  top  of  the  slide  when  the  toboggan  starts. 

44.  Derivative  of  a  Function  of  a  Function.  If  y  =  <j)(u)  and 
u  =  fix),  ?/  is  a  function  of  x.  The  derivative  of  y  with  respect 
to  X  can  be  found  without  eliminating  u.  For  any  set  of  corre- 
sponding increments,  Ax,  Ay,  and  Au, 

Ay  _  Ay  Au 
Ax      Au  Ax 
Hence 

lim  A^  ^    lim   4^    lim   Aw^ 

Since  Am  approaches  zero  as  Ax  approaches  zero, 

dx      du  dx  ^  ' 

This  is  the  formula  for  the  derivative  of  a  function  of  a  function. 
Illustration.     Let 

y  =  u^  -\-  5 
and 

u  =  3x^  +  7x  +  10. 

du 
and 


§46J  ALGEBRAIC  FUNCTIONS  67 

Then 

^  =  3m2(6x  +  7) 

=  3(3a;2  +  7x  +  10)^  (6x  +  7). 

dv 
45.  Inverse  Functions.     If  x  =  't>iv),li^  can  be  found  by  the  rule 

^  =  A, 

dx       dx 

dy 
which  is  easily  proved. 

dy  ^    Mm  ^  ^   Mm    i_  ^  ±, 
dx       '^^-0  Ax        ^x^^Ax       dx 
Ay       dy 
Illustration.     U  x  =  5y^  -\-  7y^  +  3, 

^^  =  302/«  +  147/, 


and 


dy 

dy 1 

dx      2y{lby*  +  7) 


Exercises 


1.  Find-TT-  in  terms  of  X  if: 
ax 


(a)  y  =  Vu^+  7  and  m  =  3a;  +  10. 
(6)  7/  =  2«3  +  5m  and  i*  =  x"  -  2x. 

(c)  y  =  —  and  w  =  x^  —  2. 


2.  Find  ^  if 
dx 


W    ^    -    (7/2+2)2 


3.  Find  ^  if: 
dx 


(a)  y*  +x*  -  7xy  =  15.         (6)  Sxy^  +  &x'y  +  4x2  =  15. 
46.  Parametric  Equations.     If  the  equation  of  a  curve  is  given 
in  parametric  form,  x  =  f{t),  y  =  (f>{t),  it  is  important  to  be  able 
to  find  the  derivative  of  y  with  respect  to  x  without  eliminating « 


68 


CALCULUS 


[§47 


between  the  given  equations.     A  rule  for  doing  this  can  be  derived 
by  the  method  used  in  §§13  and  17. 

If  I  is  given  an  increment  At,  x  and  y  take  on  the  increments 
Ax  and  Ay,  respectively.    Then 

Ay 
Ay      At 


and 


or 


Ax 

Ax 

At 

Um  At/ 

lim  ^y 

At=0  ^t 

M=0  ^x 

lim  Ax 

At  =  0    ^{ 

dy 

dy 
dx 

dx 

dt 

Exercises 

1.  Find  the  slope  at  (6,1)  of  the  curve  whose  parametric  equations 
arc 

X  =  t^  +t, 

2/  =  <  -  1. 
Find  -T-  for  each  of  the  following: 

2.  x  =  t", 

2/  =  i^  +  1. 

3.  x  =  u^  +3, 
1 

47.  Lengths  of  Tangent,  Normal,  Subtangent,  and  Subnormal. 

In  Fig.  31,  PT  is  the  tangent  and  FN  is  the  normal  at  P.  The 
lengths  of  the  lines  PT,  PN,  TD,  and  DN  are  called  the  tangent, 
the  normal,  the  subtangent,  and  the  subnormal,  respectively,  for 
the  point  P.     Show  that  the  lengths  of  these  lines  are; 

y^ 

¥  (1) 

dx 


y 


TD  = 


§47]  ALGEBRAIC  FUNCTIONS  69 

DN  =  y'£.  (2) 


"^-UHir- 


(3) 


dx 


=4- (I) 


PN  =  yjl  -h  (^]  •  (4) 


Exercises 

1.  Obtain  the  length  of  the  tangent,  normal,  subtangent,  and  sub- 
normal for  the  point  (1,  2)  on  the  curve  y*  =  4x.  Show  that  for  points 
on  this  curve  the  subnormal  is  of  constant  length. 
•  2.  Write  the  equation  of  the  tangent  to  y^  =  4x  at  the  pomt  (1,2). 
Write  the  equation  of  the  normal  at  the  same  point.  It  is  to  be  noted 
that  in  this  exercise  the  equations  of  the  tangent  and  normal  lines 
are  to  be  found,  and  not  the  lengths  of  the  tangent  and  normal  as 
in  the  preceding  exercise. 

3.  Write  the  equation  of  the  tangent  to 

I*      «* 

—  +—  =  1 

25  ^9        ^ 

at  the  point  (3,  2.4).     Use  implicit  differentiation. 

4.  Find  the  equation  of  the  curve  whose  subnormal  is  of  constant 
length  4  and  which  passes  through  the  point  (1,  3). 

2 
6.  Find  the  length  of  the  tangent  to  j/  =  -  at  the   point  where 

y  =  1. 

6.  Find  the  length  of  the  normal  to  the  curve  y  ■ 
point  where  x  =  3. 

7.  Find  the  equation  of  the  curve  passing  through  the  point  (1,  3) 
and  having  a  subtangent  equal  to  the  square  of  the  ordinate. 


CHAPTER  V 

SECOND  DERIVATIVE.     POINT  OF  INFLECTION 

48.  Second  Derivative,  Concavity.  Since  the  first  derivative 
of  a  function  of  x  is  itself  a  function  of  x,  we  can  take  the  deriva- 
tive of  the  first  derivative.  The  derivative  of  the  first  derivative 
is  called  the  second  derivotive.     In  the  case  of  a  function  y  of  x,  it 

di^) 
is  denoted  by  the  symbol  — -r- — ,  or  -r-  vj') ,  or  more  commonly  by 

dh/      dhi 

-j-j*     -r-^  is  read  "the  second  derivative  of  y  with  respect  to  x."    Ifr 

d^ 
must  here  again  be  remembered  that  -r-^  is  not  a  fraction  with  a 

numerator  and  a  denominator,  but  is  only  a  symbol  representing 
the  derivative  of  the  first  derivative. 

If  y  =  f{x),  the  first  derivative  of  y  with  respect  to  x  is  some- 
times written  y'  and  very  commonly  fix).  Similarly  the  second 
derivative  is  indicated  by/"(a;). 

The  derivative  of  the  second  derivative  is  called  the  third  deriva- 

d^v 
tive.     It  is  designated  by  -r-^»  or  ii  y  =  f(x),  by  f"'{x).     The  nth 

derivative  is  designated  by-i-^'  or  by/^"^^(x). 

Between  the  points  A  and  C,  Fig.  16,  where  the  curve  is  con- 
cave downward,  the  slope  of  the  tangent  decreases  from  large 
positive  values  near  A  to  negative  values  near  C.  This  means 
that  the  tangent  revolves  in  a  clockwise  direction  as  the  point  of 
tangency  moves  along  the  curve  from  A  toward  C.  Clearly  this 
will  always  happen  for  any  portion  of  a  curve  that  is  concave 
downward.  (See  Fig.  32,  a,  h,  and  c.)  The  slope  decreases  as  the 
point  of  tangency  moves  to  the  right. 

On  the  other  hand,  if  a  portion  of  a  curve  is  concave  upward, 
the  slope  of  the  tangent  increases  as  the  point  of  tangency  moves 
to  the  right.  Thus  in  Fig.  16  the  slope  of  the  tangent  is  negative 
at  C  and  increases  steadily  to  positive  values  at  E.    The  same 

70 


§49]  SECOND  DERIVATIVE  71 

thing  is  evidently  true  for  any  portion  of  a  curve  that  is  concave 
upward.  In  this  case  the  tangent  line  revolves  in  a  counter- 
clockwise direction. 

Since  the  first  derivative  of  a  function  is  equal  to  the  slope  of 
the  tangent  to  the  curve  representing  the  function,  what  has  just 
been  said  can  be  stated  concisely  as  follows: 

If  an  arc  of  curve  is  concave  upward  the  first  derivative  is  an 
increasing  function,  while  if  the  curve  is  concave  downward,  the 
first  derivative  is  a  decreasing  function. 

If  the  second  derivative  of  a  function  is  positive  between  cer- 
tain values  of  the  independent  variable  x,  the  first  derivative  is  an 
increasing  function,  the  tangent  line  revolves  in  a  counter- clock- 


FiG.  32. 

"wise  direction,  and  consequently  the  curve  representing  the 
function  is  concave  upward  between  the  values  of  z  in  question. 
If  the  second  derivative  is  negative,  the  first  derivative  is  a  de- 
creasing function  and  the  curve  is  concave  downward.  Thus  in 
Fig.  16  the  second  derivative  is  negative  between  A  and  C,  and 
between  E  and  G.  It  is  positive  between  C  and  E,  and  between 
G  and  /. 

49.  Points  of  Inflection.  Points  at  which  a  curve  ceases  to  be 
concave  downward  and  becomes  concave  upward,  or  vice  versa, 
are  called  'points  of  inflection. 

At  such  points  the  second  derivative  changes  sign.  C,  E,  and 
G,  Fig.  16,  are  points  of  inflection.  At  C,  for  instance,  the  second 
derivative  changes  from  negative  values  to  positive  values. 

Illustration  1.  Study  the  curve  y  =  |x'  by  means  of  its 
derivatives. 

Differentiating, 

dx~  ^^* 

d^y  _ 


72 


CALCULUS 


[§49 


When  a;  <  0,  -^  <  0,  -7-  =  ^x'^  is  a  decreasing  function,  and 

the    curve    y  =  |x'     is     concave    downward.     When    x  >  0, 

d^y  dy  .  . 

-Y~2  ^  ^>  TT   ^^  ^^  increasing  function,  and  the  curve  y  =  Ix^  is 

concave  upward. 


Fig.  33. 


d'^y 

At  the  point  where  x  =  0,  -7^2  changes  sign  from  negative  to 
positive,  and  the  curve  changes  from  being  concave  downward  to 


SECOND  DERIVATIVE 


73 


being  concave  upward 
inflection. 
dy 


Hence  the  point  (0,  0)  is  a  point  of 


Since  ^-  is  positive  except  when  x  =  0,    ?/ 


iX*  IS    an    in- 


creasing function  excepting  when  x  =  0.     When  x  =  0  the  curve 
has  a  horizontal  tangent. 


Fio.  34. 


In  Fig.  33  the  graphs  of  the  functions  y=  \x^  and  of  its  first 
and  second  derivatives  are  drawn  on  the  same  axes.  Trace  out 
in  this  figure  all  the  properties  mentioned  in  the  discussion. 


Illustration  2. 

Let 

y  = 

Ix' 

—  x^  -\-  Ix 

+  2. 

Differentiating, 

J 

dy  _ 
dx 

Ix- 

-2x  +  l 

d'y 
.dx^ 

h(x 
X  — 

-  l){x  - 
2. 

3). 

74 


CALCXJLUS 


[§49 


At  X  =  2,  -T^  changes  sign  from  minus  to  plus.     Hence  the  curve 

is  concave  downward  to  the  left,  and  concave  upward  to  the  right 
of  the  line  x  =  2.  The  point  on  the  curve  whose  abscissa  is  2  is 
then  a  point  of  inflection.  The  value  of  the  function  corresponding 
to  X  =  1  is  a  maximum  value,  and  the  value  of  the  function 
corresponding  to  x  =  3  is  a  minimum  value.  See  Fig.  34  for  a 
sketch  of  the  function  and  its  first  and  second  derivatives.  Trace 
out  in  the  figure  what  has  been  given  in  the  discussion. 

The  more  important  properties  of  the  function  are  put  in  tabular 
form  below. 


X 

dx^ 

dy 
dx 

Curve 

X  <  2 
X  >  2 
X  =  2 

+ 
0 

Decreasing 
Increasing 

Concave  downward. 

Concave  upward. 

Point  of  inflection  {y  =  2^), 

X  <  1 
1  <  X  <  3 
X  >  3 
X  =  1 
X  =  3 

+ 

+ 
0 
0 

Increasing, 

Decreasing. 

Increasing, 

Maximum  point  (v  =  2f). 

Minimum  point  {y  =  2), 

Exercises 

Find  the  maximum  and  minimum  points  and  points  of  inflection  of 
the  following  curves.     Sketch  the  curves, 

1.  y  =  x'  —  3x^ 

2.  y  =  x'  4-  3x2, 

3.  2/  =  2x3  +  3x2  4.  ga;  _|_  1, 

4.  2/  =  3x*  -  4x3  -  1. 
6.  y  =  x'. 

6.  t/  =  2x<  -  4x3  -  9x2  _^  27x  +  2, 
1.  y  =  6x^  -  4x3  _|_  I 


CHAPTER  VI 


APPLICATIONS 


60.  Area  under  a  Curve:  Rectangular  Coordinates.  An  im- 
portant application  of  the  anti-derivative  is  that  of  finding  the 
area  under  a  plane  curve. 

Let  APQB,  Fig.  35,  be  a  continuous  curve  between  the  ordinates 
X  =  a  and  x  =  b.  Further,  between  these  limits,  let  the  curve 
lie  entirely  above  the  X-axis.  Our  problem  is  to  find  the  area, 
A,  bounded  by  the  curve,  the  X-axis,  and  the  ordinates  x  =  a 
and  X  =  h. 


The  area  can  be  thought  of  as  generated  by  a  moving  ordinate 
starting  from  x  =  a  and  moving  to  the  right  to  a  position  DP  where 
the  abscissa  is  x.  This  ordinate  sweeps  out  the  variable  area  u, 
which  becomes  the  desired  area  A  when  x  =  b.  On  moving  from 
the  position  DP  to  the  position  EQ  where  the  abscissa  is  a;  +  Ax, 
the  ordinate  to  the  curve  takes  on  an  increment  Ay  and  the  area  u 
an  increment  Aw.  By  taking  Ax  small  enough  the  curve  is  either 
ascending  or  descending  at  all  points  between  P  and  Q.  It 
follows  at  once  from  the  figure  that 

yAx  <  Au  <  (y  -{-Ay)Ax,  (1) 


or 


Au  ,    . 

y  <x:;.<y  +  ^y- 


Ax 


75 


76  CALCUI.US  [§50 

(If  the  curve  descends  between  P  and  Q  the  signs  of  inequahty  in 

(1)  are  reversed.    The  argument  which  follows  will  not  be  affected.) 

As    Ax    approaches   zero,    Ay   approaches   zero   and   y  -\-  Ay 

approaches  y.     Hence  v-'   which   lies  between   y  and  y  +  Ay, 

approaches  y.     Thus 

lim  Am  ^ 

Aa:=0  J>^x  ^' 


or 


If  the  equation  of  the  curve  is  y  =  f(x), 

g  =  m.  (3) 

Let  F{x)  be  a  function  whose  derivative  is  /(x) .     Then 

u  =  Fix)  +  C. 

C  is  determined  by  the  condition  that  u  =  0  when  x  =  a.     Then 

C  =  -  F(a) 
and 

u  =  F{x)  -  F(a),  (4) 

an  expression  for  the  variable  area  measured  from  the  ordinate 
X  =  a  to  the  variable  ordinate  whose  abscissa  is  x.  A,  the  area 
sought,  is  obtained  by  putting  x  =  6  in  equation  (4). 

A  =  F(b)  -  F(a)  (5) 

Illustration.     Find  the  area  A  bounded  hy  y  =  x',  the  X-axis, 
and  the  ordinates  x  =  2  and  x  =  4. 


du 
dx 

=  x\ 

u 

=  \X' 

'  +  c. 

When  X 

=  2, 

u 

=  0, 

andC 

=  - 

-    A 
3* 

Then 

and 

u  = 

\x' 

_    8 
i) 

§51j  APPLICATIONS  77 

Exercises 

1.  Find  the  area  bounded  by  the  X-axis,  the  lines  x  =  1  and  x  =  2, 
and 

(a)  y  =  mx. 

(b)  y  =  x^. 

(c)  y  =  2x2  +  3x  +  1. 

2.  Find    the    area    between    the    curves    y  =  x^    and    x  =  j/'; 
.  y*  =  a{a  —  x)  and  y  =  a  —  x;  y^  =  ^x  and  y  =  2x;  y^  =  x*  and  ?/  =  x. 

3.  Find  the  area  bounded  hy  y  =  ■%/ x  +  1,  the  Z-axis,  and  the 
ordinates  x  =  0  and  x  =  2. 

51.  Work  Done  by  a  Variable  Force.  In  this  section  there  is 
given  a  method  of  finding  the  work  done  by  a  variable  force 
whose  line  of  action  remains  unchanged. 

Illustrations  of  such  variable  forces  are: 

1.  The  force  of  attraction  between  two  masses,  m  and  ilf,  is 

given  by  the  Newtonian  law 

ItMrn 
fis)  =  -^' 

where  s  is  the  distance  between  the  masses  and  A;  is  a  factor  of 
proportionality.     Note  that  the  equation  is  of  the  form 

f(s)  -  "-,■ 

2.  The  force  exerted  by  the  enclosed  steam  on  the  piston  of  a 
steam  engine  is,  after  cut-off,  a  function  of  the  distance  of  the 
piston  from  one  end  of  the  cylinder. 

3.  The  force  necessary  to  stretch  a  bar  is  a  function  of  the 
elongation  of  the  bar. 

Let  AB,  Fig.  36,  represent  a  bar  of  length  I,  held  fast  at  the 
left  end,  A.  A  force  /  is  applied  at  its  right  end  and  the  bar 
is  stretched.  It  is  shown  experimentally  that  up  to  a  certain  limit 
the  elongation,  s,  is  proportional  to  the  force  applied  (Hooke's 
Law),  i.e., 

f  =  ks, 

where  A;  is  a  constant  depending  upon  the  length  of  the  bar, 
its  cross  section,  and  the  material  of  the  bar. 


78 


CALCULUS 


[§51 


The  work  done  by  a  constant  force  in  producing  a  certain 
displacement  of  its  point  of  application  in  its  line  of  action  is 
defined  as  the  product  of  the  force  by  the  displacement.  In 
the  problem  which  we  are  considering  the  force  varies  with  the 
displacement.  The  work  cannot  be  found  by  multiplying  the 
displacement  by  the  force.  Instead  it  will  be  found  by  integrat- 
ing an  expression  for  the  derivative  of  the  work  with  respect 
to  the  displacement. 

Let  w  denote  the  work  done  in  producing  the  displacement  from 
s  =  a  to  a  variable  position  s  =  s.  Let  Aw  denote  the  work 
done  in  producing  the  additional  displacement  As.  Let  /  denote 
the  force  acting  at  s,  and  /  +  A/  the  force  acting  at  s  +  As.     A/ 


I    I 


FiG.  36. 


may  be  positive  or  negative  according  as  the  force  increases  or 
decreases  with  distance.     For  definiteness  suppose  A/  positive. 

In  producing  the  displacement  As  the  force  varies  from  /  to 
/  +  A/,  and  hence  the  work  Aw;  lies  between /As  and  (/  +  A/)  As, 
which  represent  the  work  which  would  have  been  done  had  the 
forces  /  and  /  +  A/,  respectively,  acted  through  the  distance  As. 

Hence 

/As  <  Aw;  <  (/  +  A/)  As, 


or 


f<fs<f+^f- 


As  As  approaches  zero,  A/  approaches  zero,  and  we  obtain, 

dw 


ds 


=  f. 


(1) 


§51]  APPLICATIONS  79 

Integration  gives 

w  =  F{s)  +  C,  (2) 

where  F{s)  is  a  function  whose  derivative  is  /. 

When  s  =  a,  w  =  Q,     (2)  gives  C  =  —  F{a)  and, 

w  =F{s)  -Fia).  (3) 

This  represents  the  work  done  in  the  displacement  from  s  =  a 
to  s  =  s.  The  work,  W,  done  in  the  displacement  from  s  =  a 
to  s  =  &  is  obtained  by  substituting  6  for  s  in  (3). 

W  =  F(b)  -  F(a).  (4) 

Illustration  1.  Find  the  work  done  in  stretching  a  spring  from  a 
length  of  20  inches  to  a  length  of  22  inches,  if  the  length  of  the 
spring  is  18  inches  when  no  force  is  applied  and  if  a  force  of  30 
pounds  is  necessary  to  stretch  it  from  a  length  of  18  inches  to  a 
length  of  19  inches. 

Denote  the  elongation  of  the  spring  by  s.  In  accordance  with 
Hooke's  Law, 

/  =  ks. 

Since  s  =  1  when  /  =  30,  A;  =  30  and/  =  308 
Substituting  in  equation  (1), 

-r  =  30s. 
as 

w  =  15s2  +  C. 

The  problem  is  to  find  the  work  done  in  changing  the  elonga- 
tion from  s  =  2  to  s  =  4.  When  s  =  2,w  =  0.  Hence  C  =  —  60, 
and 

w  =  15s2  -  60. 

The  required  work,  W,  is  found  by  giving  to  s  the  value  4. 

TF  =  240  -  60  =  180. 

Thus  the  work  done  is  180  inch-pounds,  or  15  foot  p  junds. 

Illustration  2.  Two  masses  M  and  m  are  supposed  concen- 
trated at  the  points  A  and  B,  respectively.    Find  the  work  done 


80 


CALCULUS 


[§51 


against  the  force  of  attraction  in  moving  the  mass  m  along  the  line 
AB  from  a  distance  a  to  a  distance  b  from  the  mass  M,  the  latter 
mass  being  fixed. 

If  /  is  the  force  of  attraction  between  the  two  masses, 

_  kmM 

f    -         g2 

Then 

dw  _  kmM 
Is  ~~     s2 
From  which 

kmM 


w  =  — 


+  C. 


-S  +AS- 


FiG.  37. 


When  s  =  a,  w  =  0.     Hence 

kmM 

and 

w  =  kmM  — 
\_a 

1- 

To  find  the  required  work,  PF,  let  s  =  6, 

W  =  kmM  -  - 
[a 

1- 
b_ 

Illustration  3.  Gas  is  enclosed  in  a  cylinder,  one  end  of  which  is 
closed  by  a  movable  piston.  Find  the  work  done  by  the  gas  in 
expanding  in  accordance  with  the  law  pv^-*  =  K,  from  a  volume 
of  3  cubic  feet  at  a  pressure  of  15,000  pounds  per  square  foot  to 
a  volume  of  4  cubic  feet. 

Let  A  be  the  area  of  the  cross  section  of  the  cylinder.  Then 
pA  is  the  force  on  the  piston,  Fig.  37,  and 

dw 
ds=P^' 


§51] 

APPLICATIONS 

or 

dw 

AK         AK           K      I 

ds 

~  v^*  ~  (AsY*  ~  A'^U^*' 

Integration  i 

gives 

1     K     1 

^~       0.4  ^""s"'*'^^ 

-        ^    ^  +C 

When  V  =  3 

,w 

=  0. 

Hence 

1    K 
0.4  3°"' 

and 

K  r  1         1  -] 
^     0.4  [s^*     v°*y 

0.4  [3°"      40''_ 

81 


When  t;  =  3,  p  =  15,000.    Hence  K  =  (15,000)(3i"),  and 

W  =  150,000[0.75  -  (0.75)  i«l 
=  12,230. 

Exercises 

1.  A  spring  is  12  inches  long  and  a  force  of  120  pounds  is  necessary 
to  stretch  it  from  its  original  length,  12  inches,  to  a  length  of  14  inches. 
Find  the  work  done  in  stretching  the  spring  from  a  length  of  13  inches 
to  a  length  of  15  inches. 

2.  In  the  case  of  a  bar  under  tension,  Fig.  36,  the  relation  between 
the  stretching  force,  /,  the  original  length  of  the  bar,  I,  and  the  elonga- 
tion of  the  bar,  s,  is  given  by 

^       EAs 

^  =  -r' 

where  E  is  the  modulus  of  elasticity  of  the  material  of  the  bar  and 
A  is  the  area  of  the  cross  section  of  the  bar.  Find  the  work  done  in 
stretching  a  round  iron  rod  ^  inch  in  diameter  and  4^  feet  long  to  a 
length  of  54.5  inches,  given  that  E  =  3-10^  pounds  per  square  inch. 

3.  A  spherical  conductor,  A,  is  charged  with  positive  electricity  and 
a  second  spherical  conductor,  B,  with  negative  electricity.  The  force 
of  attraction  between  them  varies  inversely  as  the  square  of  the 

6 


82 


CALCULUS 


[§52 


distance  between  their  centers.  If  the  force  is  10  dynes  when  the 
centers  are  100  centimeters  apart,  find  the  work  done  by  the  force  of 
attraction  in  changing  the  distance  between  the  centers  from  140 
centimeters  to  120  centimeters. 

4.  Find  the  work  done  by  a  gas  in  expanding  in  accordance  with 
the  law  pv^  =  C  from  a  volume  of  5  cubic  feet  to  one  of  6  cubic 
feet,  if  p  =  70  pounds  per  square  inch  when  v  =  5  cubic  feet. 

5.  Find  the  work  done  in  compressing  a  spring  6  inches  long  to  a 
length  of  5§  inches  if  a  force  of  2000  pounds  is  necessary  to  compress 
it  to  a  length  of  5  inches. 

6.  The  work  done  by  a  variable  force  can  be  represented  graphically 
as  the  area  under  a  curve  whose  ordinates  represent  the  force.  Con- 
struct the  figures  and  prove  this  fact  for  Illustrations  1,  2,  and  3. 

62.  Parabolic  Cable.  Suppose  a  cable,  AOB,  Fig.  38,  a,  is 
loaded  uniformly  and  continuously  along  the  horizontal,  i.e.,  so 
that  any  segment  of  the  cable  sustains  a  weight  proportional  to 


the  projection  of  the  segment  upon  a  horizontal  line.  Let  k  be  the 
weight  carried  by  a  portion  of  the  cable  whose  horizontal  pro- 
jection is  one  unit  of  length. 

Choose  0,  the  lowest  point  of  the  cable,  as  origin  and  a  hori- 
zontal line  through  0  as  axis  of  x.  Let  P  be  any  point  on  the  cable. 
Suppose  the  portion  OP  of  the  cable  cut  free,  Fig.  38,  b.  To  keep 
this  portion  in  equilibrium  a  horizontal  force  H  and  an  inclined 
force  T  must  be  introduced  at  the  points  0  and  P,  respectively. 
The  force  H  must  be  equal  in  magnitude  to  the  tension  in  the  cable 
at  0,  and  it  must  act  in  the  direction  of  the  tangent  line  at  that 
point.  Similarly,  the  force  T  must  be  equal  to  the  tension  in 
the  cable  at  the  point  P  and  act  in  the  direction  of  the  tangent 
line.     The  force  T  can  be  resolved  into  its  vertical  and  horizontal 


§53]  APPLICATIONS  83 

components  V  and  H',  respectively.  Now  H  and  H'  are  the  only 
horizontal  components  of  the  forces  acting  on  OP  and,  since  OP  is 
in  equilibrium,  they  must  balance  each  other.     Therefore, 

H  =  H'.  (1) 

Hence  the  horizontal  component  of  the  tension  in  the  cable  is 

independent  of  the  point  P,  i.e.,  it  is  a  constant. 

In  like  manner  the  only  vertical  components  of  the  forces 

acting  on  OP  are  the  weight  kx  supported  by  OP,  acting  downward, 

and  V,  the  vertical  component  of  T.    They  must  balance  one 

another.     Hence 

V  =  kx.  (2) 

V 
The  slope  of  the  tangent  Ime  to  the  curve  at  the  point  P  is  „/ 

Then 

dy  _   V^  _kx 

dx~  H'~  H'  ^'^' 

This  is  the  slope  of  the  curve  at  any  point.  On  integrating 
we  obtain  the  equation  of  the  curve  apart  from  the  arbitrary 
constant  C. 

y  =  %  +  c.  (4) 

C  is  determined  by  the  condition  that  y  =  0  when  x  =  0.  Then 
C  =  0,  and  (4)  becomes 

y  =  2^-  (5) 

This  is  the  equation  of  a  parabola  with  its  vertex  at  the  origin. 

63.  Acceleration.^  In  §38  acceleration  was  defined  as  the  time 
rate  of  change  of  velocity,  i.e.,  as  the  derivative  of  the  velocity 
with  respect  to  the  time.  But  velocity  is  the  derivative  of  dis- 
tance with  respect  to  time.  Hence  the  acceleration  is  the  second, 
derivative  of  the  distance  with  respect  to  the  time.     If  s  denotes 

the  distance  and  t  the  time,  the  acceleration  is  expressed  by  -w-. 
In  the  case  of  a  freely  falling  body 

d'^s  „. 

•  The  statements  in  this  section  refer  to  motion  in  a  straight  line. 


84  CAIiCULUS  [§53 

The  relation  between  s  and  t  can  be  found  from  this  differential 
equation  by  integrating  twice,  as  follows: 
The  first  integration  gives 

and  the  second 

s  =  igt^  +  Cit  +  C2.  (3) 

Two  arbitrary  constants  of  integration  are  introduced.     They 
can  be  determined  by  two  conditions.     If 

s  =  So  (4) 

and 

v  =  -^  =  Vo  (5) 

when  t  =  0,  (2)  gives  Ci  =  Vq,  and  (3)  gives  C2  =  Sq.     Then 

s  =  hgt'  +  vot  +  So.  (6) 

This  result  was  found  in  §38  by  essentially  the  same  method,  where 

dv  d^s 

the  symbol  n^  was  used  instead  of  -n^' 

Exercises 

1.  Solve  Exercise  5,  §38,  by  the  method  used  above. 

2.  Obtain   the   relation    v  =  \/2gs    (see    §38)    directly   from    the 
equation 


ds 
Hint.     Multiply  by  2  -jr  • 


dt^ 


=  9- 


ds  d^s   _  n    f^s 
^dt  di^  ~^^dt' 

(ds\i 
~T7  I    with  respect  to  t  and  the 

second  that  of  2gs.     We  then  have 

Determine  C  by  the  condition  that  v  =  Q  when  8  =  0. 


§54] 


APPLICATIONS 


85 


64.  The  Path  of  a  Projectile.  An  interesting  application  of 
integration  is  to  find  the  equation  of  the  path  of  a  projectile,  a 
baseball  for  instance,  thrown  with  a  given  velocity  at  a  given 
inclination  to  the  horizontal. 

Let  0,  Fig.  39,  be  the  point  from  which  the  ball  is  thrown. 
Take  this  point  as  the  origin  of  a  system  of  rectangular  coordi- 
nates. Let  the  ball  be  thrown  so  that  its  direction  at  the  instant 
of  leaving  the  hand  makes  an  angle  a  with  the  horizontal,  and  let 
the  initial  velocity  of  the  ball  be  Vq.  Then  the  horizontal  com- 
ponent of  the  initial  velocity  is  Vo  cos  a,  while  its  vertical 
component  is  Vo  sin  a.  That  is,  at  the  instant  the  ball  is  thrown 
its  x-coordinate  is  increasing  at  the  rate  of  Vo  cos  a  feet  per  second. 
Similarly  the  initial  rate  of  change  of  the  ^/-coordinate  is  vq  sin  a. 


Fig.  39. 


At  the  end  of  t  seconds  after  the  ball  was  thrown  it  is  at  the  point 
P  whose  coordinates  are  x  and  y.  If  the  resistance  of  the  air  is 
neglected  there  is  no  force  acting  on  the  ball  tending  to  change  the 
component  of  its  velocity  parallel  to  the  X-axis.  Hence  the 
x-component  of  the  velocity  is  at  all  times  the  same  as  at  the 
beginning,  viz.,  Vq  cos  a.     The  x-component  of  the  velocity  is  also 

dx 

-rr,  viz.,  the  time  rate  of  change  of  the  abscissa  of  the  ball.     There- 
fore we  can  write 

dx  .  . 

jT  =  Vo  cos  a.  {1) 

From  which  on  integration 

X  =  {vo  cos  a)t  +  C.  (2) 

Time  is  counted  from  the  instant  the  ball  was  thrown.     The 


86  CALCULUS  [§54 

condition  for  determining  C  is  then  that  a;  =  0  when  f  =  0.  It 
follows  that  C  =  0,  and  (2)  becomes 

X  =  {vo  cos  a)t.  (3) 

This  equation  gives  the  x-coordinate  of  the  ball  at  any  time  t. 

In  the  vertical  direction,  the  force  of  gravity  acts  to  change  the 
t/-component  of  the  velocity. 
Then 

g=    -..  (4) 

The  negative  sign  is  used  since  the  force  of  gravity  causes  the 
velocity  in  the  direction  of  the  positive  F-axis  to  decrease.  In- 
tegration gives 

1=   -gl  +  C.  (6) 

dv 
d  is  determined  by  the  condition  that  37  =  ^0  sin  a  when  t  =  0. 

Then  C2  =  Vq  sin  a  and  (5)  becomes 

^y  =    —  gt  +  v^  sin  a.  (6) 

Integrating  again, 

y  =    -  W  +  i^o  sin  ci)t  +  C3.  (7) 

Since  y  —  0  when  ^  =  0,  C3  =  0,  and  (7)  becomes 

y  =    —  hgi^  +  (^0  sin  a)L  (8) 

This  is  the  ^/-coordinate  of  the  ball  at  any  time  t.  Equations  (3) 
and  (8)  are  the  parametric  equations  of  the  path  of  the  ball. 
The  elimination  of  t  between  these  equations  gives  the  rectangular 
equation  of  the  path, 

y  =    —  :r~2 r~  x^  +  x  tan  a.  (9) 

"  2^0   cos^  a 

This  is  the  equation  of  a  parabola  with  its  vertex  at  the  point, 

r?;o^  sin  2a     Vq'^  sin^  a"! 
L       2g       '   ~2^J" 

Tt  is  to  be  remembered  that  in  the  solution  of  this  problem  the 


§54]  APPLICATIONS  87 

resistance  of  the  air  was  neglected.  Consequently  the  results 
obtained  can  be  regarded  only  as  approximations.  Experimentally 
it  has  been  shown  that  the  resistance  of  the  air  increases  with  the 
velocity  of  the  moving  body.  For  low  velocities  the  resistance  is 
assumed  to  vary  as  the  first  power  of  the  velocity,  but  for  higher 
velocities,  such  as  are  attained  by  rifle  balls,  the  resistance  is  as- 
sumed to  vary  with  the  second  power,  and  the  results  obtained 
above  cannot  be  considered  to  be  even  approximations. 

Exercise 

1.  Find  the  angle  of  elevation,  a,  at  which  the  ball  must  be  thrown 
to  make  the  range,  OA,  Fig.  39,  a  maximum. 


CHAPTER  VII 
INFINITESIMALS,    DIFFERENTIALS,    DEFINITE   INTEGRALS 

55.  Infinitesimals.  In  §23  an  infinitesimal  was  defined  as  a 
variable  which  approaches  the  limit  zero.  Thus,  x^  —  1,  as  z 
approaches  1,  is  an  infinitesimal. 

It  is  to  be  noted  that  a  variable  is  thought  of  as  an  infinitesimal 
only  when  it  is  in  the  state  of  approaching  zero.  Thus  x^  —  1  is 
an  infinitesimal  only  when  x  approaches  +1  or  —1.  An  in- 
finitesimal has  two  characteristic  properties:  (1)  It  is  a  variable. 
(2)  It  approaches  the  limit  zero;  i.e.,  the  conditions  of  the  problem 
are  such  that  the  numerical  value  of  the  variable  can  be  made  less 
than  any  preassigned  positive  number,  however  small. 

This  meaning  of  the  word  infinitesimal  in  mathematics  is  entirely 
different  from  its  meaning  in  everyday  speech.  When  we  say 
in  ordinary  language  that  a  quantity  is  infinitesimal,  we  mean 
that  it  is  very  small.  But  it  is  a  constant  magnitude  and  not  one 
whose  numerical  measure  can  be  made  less  than  any  preassigned 
positive  number,  however  small.  Thus,  0.000001  of  a  miUigram 
of  salt  might  be  spoken  of  as  an  infinitesimal  quantity  of  salt,  but 
the  number  0.000001  is  clearly  not  an  infinitesimal  in  the  sense 
of  the  mathematical  definition.  On  the  other  hand,  if  we  have  a 
solution  containing  a  certain  amount  of  salt  per  cubic  centimeter 
and  allow  pure  water  to  flow  into  the  vessel  containing  the  solu- 
tion while  the  solution  flows  off  through  an  overflow  pipe,  the 
quantity  of  salt  per  cubic  centimeter  constantly  diminishes.  The 
amount  of  salt  left  in  solution  after  a  time  t  is  then  an  infinitesimal, 
as  t  becomes  infinite. 

Infinitesimals  are  of  fundamental  importance  in  the  Calculus. 
The  derivative,  which  we  have  already  used  in  studying  functions, 
is  the  limit  of  the  ratio  of  two  infinitesimals.  Ay  and  Ax. 

56.  ^f^  •    Let  the  arc  AB,  Fig.  40,  subtend  an  angle  a 

88 


INFINITESIMALS 


89 


at  the  center,  O,  of  a  circle  of  radius  r.  The  angle  a  is  measured 
in  radians.  Let  AT  he  tangent  to  the  circle  at  A,  and  let  BC  be 
perpendicular  to  OA.  The  area  of  the  triangle  OCB  is  less  than 
the  area  of  the  circular  sector  OAB,  and  this  in  turn  is  less  than 
the  area  of  the  triangle  OA  T. 

UBC){OC)  <  har^  <  ^iAT)r 

BC  OC  ^       ^AT 
—  — -  <  a  <  

r      r  r 

OC   . 

■ —  sin  a  <  a  <  tan  a 
r 


OC 


< 


sin  a 

As  the  angle  a  approaches  zero, 

OC 
OC  approaches  r  and  —  approaches 

1,  and  further,  cos  a  approaches  1. 
Hence  the  first  and  last  members 
of  the  inequalities  (2)  approach  the 
same    limit,    1.    Then  the  second 

a 

member,  -; — ■»  which  lies  between 
'  sin  a 

them,  must  approach  the  same  limit,  1.     Therefore 


lim   sin  a 


lim 

a=0 


1 

a 

sin  a 


=  I. 


(3) 


57. 


lim   tan  a    lim   tana 
sin  a 

tan  a 


a=0 


lim 


a 


^  lim     /sina      1    \ 
«-o    \    a     cos  a/ 

^   Aim   sina\  /lim       1    \ 
\a=0      cc   /  \°'=°   cos  OC/ 


=  1. 


lim   tan  a 


lim 

a=0 


=  1. 


(1) 
(2) 


58. 


lira    1  —  cos  a 


1=0 


90  CALCULUS                                          [§59 

Since 

^  2  sin''  fT      sin  7, 

1  —  cos  a  2              2    .     a 


=  =  sin  ^» 

a  a  a  2 


lim    l-coso!       /lim   ^^"  2  W  IJm     .    a 
cc^Q  a         ~  I  «=^o      a     /  I  «^o  ^      2 

Hence 

lim   1  —  cos  «  _ 

a=^o  «         ~  "•  ^^^ 

In  Fig,  40,  AB,  AC,  AT,  BT,  and  BC  are  infinitesimals  as  a 
approaches  zero.     Then, 

from  (3),  §56,  i™   ^  =  h 

from  (1),  §57,  i^o  ^  =  1, 

from  (2),  §57,  i^o  ^^  =  1, 

from  (1),  §58,  i^o  J^  =  0. 

59.  Order  of  Infinitesimals.  Consider  the  infinitesimals  x^ 
and  X  as  X  approaches  zero.  The  ratio  of  x^  to  x  is  x,  which  is 
itself  an  infinitesimal.  The  infinitesimals  x^  and  x  are  repre- 
sented. Fig.  41,  by  the  ordinates  MP  and  MN,  to  the  curves 
y  =  x^  and  y  =  x.    The  quotient 

X   ~  MN 

is  a  measure  of  the  relative  magnitude  of  these  infinitesimals  as 
they  approach  zero.     It  shows  that  MP  becomes  small  so  much 
more  rapidly  than  MN  that  the  limit  of  their  quotient  is  zero. 
On  the  other  hand,  the  infinitesimals  2x  and  x  behave  very 

2x 
differently.     Their    quotient   is   —  =  2,   and  the  limit  of  this 


J59] 


INFINITESIMALS 


91 


quotient  is  2.     In  this  case  the  limit  of  the  ratio  of  the  infini- 
tesimals is  not  zero.     (See  Fig.  42.) 
Again, 

lim    1  —  cos  a 

while 


=  0, 


lim   sma 

«-0         rv 


=   1. 


These  illustrations  of  the  comparison  of  two  infinitesimals  lead 
to  the  following  definitions  of  the  order  of  one  infinitesimal  with 
respect  to  another. 

Two  infinitesimals,  a  and  /3,  are  said  to  be  of  the  same  order  if  the 

ex 

limit  of  ^  is  a  finite  number  not  zero. 


Fig.  41. 


Fig.  42. 


7/  the  limit  of  ^  is  zero^  a  is  said  to  be  of  higher  order  than  jS. 

Thus,  2x  and  x  are  of  the  same  order;  x^  is  an  infinitesimal  of 
higher  order  than  x;  sin  a  and  a,  or  CB  and  AB,  Fig.  40, 
are  of  the  same  order;  tan  a  and  a,  or  AT  a,nd  AB,  Fig.  40,  are  of 
the  same  order;  tan  a  and  sin  a,  or  CB  and  AT,  Fig.  40,  are  of  the 
same  order;  1  —  cos  a  is  of  higher  order  than  a,  or  CA,  Fig.  40, 
is  of  higher  order  than  AB. 

Let  ACB,  Fig.  43,  be  a  right  angle  inscribed  in  a  semicircle. 
Let  BD  be  a  tangent  line,  and  let  CE  be  perpendicular  to  BD.     If 


92 


CALCULUS 


[§59 


the  angle  CAB  approaches  zero,  BC,  CD,  BE,  ED,  CE,  and  arc 
BC  are  infinitesimals.     From  similar  triangles 


AB       BC      BD 

AC~  BE~  BC' 

AB 
Since  lim  a^—  1,  it  follows  that 

,.     BC      ..     BD       ^ 

Hence  BC  and  BE,  and  BD  and  BC  are  infinitesimals  of  the  same 

order. 

Again, 

BC       CE       CD 
AB      BC  ~  BD' 

D 

E 

,•    BG    ^ 

Smct!  lim  -ir-f,  =  0, 

AB         ' 


Fig.  43. 


Hence  CE  is  an  infinitesimal  of  higher  order  than  BC,  and  CD 
is  an  infinitesimal  of  higher  order  than  BD. 
Again, 

CB  ^CE^  _CD 

AC  ~  BE  ~  CB 

Since  lim  -j-^  =  0, 

AC         ' 

hm^  =  lim^  =  0. 

Hence  CE  is  an  infinitesimal  of  higher  order  than  BE,  and  CD  is 
an  infinitesimal  of  higher  order  than  CB. 


§60]  INFINITESIMALS  93 

Exercises 

1.  Show  that  X  —  2x^  and  3a;  +  x^  are  infinitesimals  of  the  same 
order  as  x  approaches  zero. 

2.  Show  that  1  —  sin  d  and  cos''  9  are  infinitesimals  of  the  same 
order  as  6  approaches  g- 

3.  Show  that  1  —  sin  0  is  an  infinitesimal  of  higher  order  than 
cos  0  as  6  approaches  ^^ 

4.  Show  that  sec  a  —  tan  a  is  an  infinitesimal  as  a  approaches  -• 

5.  Show  that  1  —  sin  a  is  an  infinitesimal  of  higher  order  than 
sec  a  —  tan  a  as  a  approaches  ^• 

6.  Show  that  1  —  cos  6  is  an  infinitesimal  of  the  same  order  as  0^ 
as  6  approaches  zero. 

T    ou       +1,  +  lim  sin  g  -  g 

7.  Show  that  ^^q =  0. 

o    ou       +1.  +  lim  sine  -  9 

8.  Show  that  „ .  „  — . — r—  =  0. 

»=o     sin  9 

9.  Show  that  i™  *^^^T^^  =  °- 

10.  Show  that  1™*^J^^  =  0. 

*=o     tan  d 

^^    n,         .  >     ,  lim  sin  a  —  tan  a 

11.  Show  that     .^ 7 =  0. 

«=0        tan  a 

12.  Show  that^'"^/^".-^^""  =  0. 

«=o        sin  a 

60.  Theorem.     The  limit  of  the  quotient  of  two  infinitesimals, 
a  and  /3,  is  not  altered  if  they  are  replaced  by  two  other  infinitesimals, 

a  B 

y  and  5,  respectively,  such  that  lim  --  =  1  and  lim  -:  =  1. 


Proof  : 


a 

y  ~ 

a  _        7 

~B~  TT' 

lim  — 

lim  ^  =  a   lim  -i  =  lim  j  i 

hm   g^ 


94  CALCULUS  [§60 

since 

lim  —   =  lim  -v  =  1- 
7  0 

It  is  evident  from  the   proof   that  the  limit  of  the  quotient  is 

unaltered  if  only  one  of  the  infinitesimals,  say  a,  is  replaced  by 

T 
another  infinitesimal  7,  such  that  lim  -  =  1. 

a 

Illustrations. 

L  Since 

lim  sin  a  _ 

a=0     ~      ~  ^» 


a 

lim  1  ~"  cos  a  _  Hm  1  ~  cos  a 
«-o     sin  a      ~  "=o         q. 


=  0. 


2.    Since 


lim  tan  a 


a  =  0 


a 


1, 


lim  1  —  cos  «  _  lim  1  —  cos  «  _ 
«-0     tan  a  "-0         a 

3.    In  Fig.  40, 

lim  ^  ^  lim  C;4  _  Urn  C'A  ^ 
cc^oAB       »=OBC       "=0AT 


Exercises 
l.Showthat^^"l^"7"  =  i 
Hint.     1  —  cos  a  =  2  sin^  ^- 

2.  Show  that  lim^i^«a--cos«)  ^ 

a=0  a^  ^ 

3.  Show  that  lim(«-5)-sina  ^  ^5. 

a=0  a 

4.  Show  that"'"    ^"'"'«         ' 


«=Ocos  a  sin-  a        ^* 
K    «,         ,,     .  lim  33:"  -  4x3 

5.  Show  that  ;,^0  2x^-5x<  =  ^' 

Hint.     Replace  numerator  by  3x*  and  denominator  by  2x*. 
A    .A  A 

6.  Show  that   "."^^—4^=    ^^-^  =  2. 

X^  X^  X^ 


1611 


DIFFERENTIALS 


95 


61.  Diflferentials.  Let  PT,  Fig.  44,  be  a  tangent  line  drawn 
to  the  curve  y  =  f{x)  at  the  point  P.  Let  DE  =  Ax,  RQ  = 
A?/,  and  let  angle  RPT  =  t. 

From  the  figure, 

RM      .  .,.  . 

=  tanr  =  /  (x), 


Ax 


or 


RM  =  fix) 'Ax. 

This  is  the  increment  which  the  function  would  take  on  if 
it  were  to  change  uniformly  at  a 
rate  equal  to  that  which  it  had 
at  P. 

This  quantity,  j'{x)Ax,  is 
called  differential  y,  and  is  de- 
noted by  dy.  Its  defining  equa- 
tion is 


dy  =  f'{x)Ax. 


(1) 


Ax,  the  increment  of  the  independent  variable,  is  called  differential 
X  and  is  denoted  by  dx,  i.e.,  Ax  =  dx.     Equation  (1)  becomes 


dy  =  j'{x)dx\ 


(2) 


In  Fig.  44,  RM  =  dy  and  DE  =  PR  =  dx. 

In  general,  dy  is  not  equal  to  Ay,  the  difference  being  MQ, 

Fig.  44.     However,  it  will  be  shown  that    ^^^g  ^ —  ~  ^• 

lim    RQ_f,(^s 


or 


lim   \RQ_  Rm  =  .>r^. 

ix=0      RM  PR  J   ^•^^' 


^^^0  IRM  PRJ 


(3) 


RM . 


But,  since  p^  is  constant  and  equal  to/'(x),  equation  (3)  becomes 
lim    RQ 

Ax=0    RM 


1, 


>  In  the  expression  (2)  for  the  differential  of  the  function /(i),  the  first  derivative 
is  the  coefficient  of  the  differential  of  the  argument,  and  for  this  reason  it  is  sometimes 
called  the  differential  coefficient. 


96  CALCULUS  (§62 

at 

It  is  to  be  noted  that  dx  is  an  arbitrary  increment  and  that  dy 
is  then  determined  by  this  increment  and  the  value  of  the  deriva- 
tive, i.e.,  by  the  slope  of  the  tangent  at  the  point  for  which  the 
differential  is  computed,  dx  and  dy  are  then  definite  quantities 
and  we  can  perform  on  them  any  algebraic  operation.  Thus  we 
can  divide  (2)  by  dx  and  obtain 

t  =  f'M.  (5) 

where  dy  and  dx  denote  the  differentials  of  y  and  x,  respectively. 
Thus  from  the  definition  of  differentials  the  first  derivative  may 
be  regarded  as  the  quotient  of  the  differential  of  y  by  the  differ- 
ential of  X. 

It  is  to  be  observed,  however,  that  this  statement  gives  no  new 
meaning  to  the  derivative,  since  the  derivative  was  used  in  the 
definition  of  the  differential. 

62.  Formulas  for  the  Differentials  of  Functions.  In  accordance 
with  equation  (5)  of  the  preceding  section,  any  formula  involving 
first  derivatives  can  be  regarded  as  a  formula  in  which  each  first 
derivative  is  replaced  by  the  quotient  of  the  corresponding 
differentials.     Thus, 

,  /m\  du         dv 


dx  dx 


dx 


Each  derivative  being  considered  as  a  fraction  whose  denominator 
is  dx,  we  can  multiply  by  dx,  and  obtain 

vdu  —  udv 


e)  = 


v 


In  words,  the  differential  of  a  fraction  is  equal  to  the  denominator 
times  the  differential  of  the  numerator  minus  the  numerator  times  the 
differential  of  the  denominator,  all  divided  by  the  square  of  the  denomi- 
nator. It  will  be  noted  that  the  wording  is  the  same  as  that 
for  the  derivative  of  a  fraction  except  that  throughout  the  word 
differential  replaces  the  word  derivative. 


§62]  DIFFERENTIALS  97 

The  other  formulas  for  derivatives  which  have  been  de- 
veloped are  expressed  below  with  the  corresponding  formulas  for 
differentials. 


Formulas 

1. 

dc 

dc  =  0. 

2. 

d{cu)         du 
dx     ~     dx 

d{cu)  =  cdu. 

3. 

d{u  +  v)       du        dv 
dx        ~  dx        dx 

d{u  +  v)  =  du  +  dv. 

4. 

du'*               ,   du 
dx                   dx 

dw*  =  nM"~^  du. 

6. 

d(uv)          dv          du 
dx            dx         dx 

d(uv)  =  udv  -{-  vdu. 

6. 

,  /u\           du          dv 

a       )         v~j wt— 

\v  1           dx          dx 

dx     ~            v^ 

,  /u\      vdu  —  udv 

7. 

\v  /                dx 

dx      ~          v^ 

,  /c  \             cdv 

8. 

dx                v*^ 

/  c  \              cndv 

9. 

du 
du          dx 

dJ  -    ^"  . 
2u^ 

The  formula  for  the  differential  oi  y  =  cu"  can  be  put  in  the  fol- 
lowing convenient  form : 

(fy  du 

10.  —  =  n — ■} 

y         u 

which  is  obtained  directly  by  dividing  dy  =  cnu'*~^du  hy  y  =  cW. 

The  process  of  finding  either  the  derivative  or  the  differential  of  a 
function  is  called  differentiation. 

The  process  of  finding  a  function  when  its  derivative  or  differential 
is  given  is  called  integration. 

We  have  no  symbol  representing  integration  when  applied  to 
derivatives.     The  symbol  for  integration  when  applied  to  dif- 


98  CALCULUS  [§62 

forentials  is    j  .     Thus    I  Sx^dx  =  x^  -\-  C.    The  origin  of  this 
symbol  will  be  explained  later.     It  is  read  "integral  of." 

Illustrations. 

1.  U  y  =  Vn^S 

dy  =  Ui  -  x2)-5(  -  2xdx) 
_  xdx 

Vl  -  x^' 
By  formula  10,  where       w  =  1  —  x^, 

dy  _  1  —  2x  dx 
J  ~2  (1  -  a;*) 

xdx 


2.  If   2/  = 


1  -a;* 

X 


X2-    1 


^y  = (^^^^Tp 

_  (x'  —  1)  c?a;  —  X  (2a;  dx) 
~  (x2_  1)2  ■ 

^  _  (x^  +  l)dx 
(x*  -  ly ' 
3.  If    d?/  =  xdx, 


y  =  ixdx 

=  A  r2xdx 


=i-+c. 


4.  If   dy  =  xVl  —  x^  dx, 

y  =    Cx{l  —  x^)^dx 
=  -H/HI  -x2)2(  _2xdx) 
(l-x^)^ 


+  C. 


§62]  DIFFERENTIALS  •  99 

dtj         dx 

r' 

2/  =  C(x  -  1). 


5.  If  -  , 

y      X  -  1 


by  formula  10. 
ft    If      d]i       xdx 


y       x^  —  1 


dy  _  1  2xdx 
y  ^  2  x2-  1 

y  =  C\/a;2  -  1. 


Exercises 

Find  dy  in  the  following  ten  exercises : 

1.  J/  =  x2  -  3x  -  2.  ^  V: 

6.  y  = 


2.  2/  = 


(x-l)» 


X-  1  7.  J/  =  (x  -  l)(x»  -  1)2. 

3.  y  =  x^-  x~^  -  3a;.  8.  y  =  {x*  +  x  -  2)». 


,_i 


4.  y  =  (x  -  2)*.  9.  y  =  (x  -  1)" 

6.  y  =  (x2  -  2)i  10.  y  =  (x2  -  1)~^. 

Integrate  the  following: 

11.  fxMx. 

12.  C{x^  -  l)xdx. 

13.  fCx'  -  3x  +  5)  (x2  -  1)  dx. 

14.  r(x»  -  2x  -  6)3  (x  -  1)  dx. 

/dx 

16.    Cs/x  dx. 

"■  /t 

18.    Cx^dx. 


100 


CALCULUS 


m 


19. 

dy 

y 

= 

X 

20. 

dy 

_ 

xdx 

x2-  1 


21. 


dy  _     (x^  -  2x  +  4)dx 
y    ~  a;3-3x2+12a;-2' 


lim 

Ax 


Since^ 


63.  Differential    of    Length    of 
Arc :     Rectangular     Coordinates. 

Let  PR,  Fig.  45,  =  Ax,  RQ  =  Ay, 
the  chord  PQ  =  Ac,  and  the  arc 
PQ  =  As.  (s  represents  the  length 
of  arc  measured  from  some  point 
A.)    PT  is  the  tangent  at  P. 

(Ax)2  +  {Ayy 

m    /^\2  _  1    ,     lim    (^Y__  1    ,    (f^vy 
^0   \Ax)         ^  "^  ^^=^0  \^xj   ~  ^  "^  \dxl 

lim    Ac 


A2:=0    ^s 


=    1, 


>  When  Ai  is  taken  so  small  that  the  curve  has  no  point  of  inflection  between  P 
and  Q,  the  chord  PQ  <  arc  PQ  <  PT  +  TQ,  or  Ac  <  As  <  Pr  +  TQ.     Whence, 


Therefore 


Then  from  (1), 


i<^<^+m 


(ptX  2  ^  (da:) 2  +  (dy)2  ^  [^  \dx) 
[ac  J  (Ax)  2  +  (A2/)2        J  _|_  /^\f 


lim     /• 
Ax=0  I 


Ac  I 


=  1. 


lim       TQ 
Ax=0    -^ 


lim       Aj/  —  dy  Ay 

^^-0     A^     a;^ 


r    lim    /.        ''i/^l  r  'im       ^ l  „  n 
=  [Ax^oV^  -  Ayjj  [ax=0    Ac  J       "' 


lim      ^  =, 
Ax— 0  Aj/ 

lim        As 
Ac 


Aa;=s=0 


(1) 


DIFFERENTIALS  101 

Ac  can  be  replaced  by  As  ( §60) . 

lira   /As\  2  ^  ,    ,  /M ' 
^=0  \Ax)  "^  [dxj  ' 


or 


(ds)2  =  (dx)2  +  (dy)2  (1) 


ds=Vl+(g)'dx  (2) 


ds  =  Vi  +  a~)^<iy 


Vdy/    -  (3) 

Equation  (1)  shows  that  the  line  PT,  Fig.  45.  represents  ds. 
If  T  denotes  the  angle  made  by  the  line  PT  with  the  positive 
X-axis, 

dx  =  cos  T  ds 

dy  =  sin  r  ds. 

Illustration.     Find  the  length  of  the  curve  y  =  fx^  between 
the  points  whose  abscissas  are  3  and  8. 

^=  x^ 
dx 


(I)'- 


Substituting  in  formula  (2), 


ds  =  Vl  +  X  dx. 
Integrating, 

s=  1(1 +  x)'  +C. 
When  X  =  3,  s  =  0.     Hence  C  =  -  Y,  and 


s=  iil  +  xy  -W 

This  formula  gives  the  length  of  the  curve  measured  from  the 
point  whose  abscissa  is  3  to  the  point  whose  abscissa  is  x.  On 
placing  X  =  8  we  obtain  s  =  ^3^,  the  length  of  the  curve  from 


102 


CALCULUS 


I  §64 


the  point  corresponding  to  x  =  3  to  the  point  corresponding  to 
X  =  8. 

Exercises 
Find  the  differentials  of  the  length  of  the  following  curves: 


1.  y  =  x'. 

2.  x^  +  y^  =  4. 

3.  2/  =  xK 

4.  2/2  =  X. 


5.  3x2  +  4y2  =  12. 

6.  xy  =  1. 

7.  xj/2  =  1. 

8.  7/  =  x-^ 


64.  The  Limit  of  2f(x)Ax.  Let  y  =  /(x)  be  a  continuous 
function  between  x  =  a  and  x  =  6.  In  §50  it  was  shown  that 
the  area  bounded  by  the  curve,  the  X-axis,  and  the  ordinates 
X  =  a  and  x  =  6  is  given  by  the  formula 

A  =  F{b)  -  F{a),  (1) 

where  F{x)  =  J  f(x)dx.  A  second  expression  will  now  be  found 
for  the  area.  Divide  the  interval  h  —  a,  Fig.  46,  into  n  equal  parts 
and  at  each  point  of  division  erect  an  ordinate.  Complete  the 
rectangles  as  indicated  in  the  figure. 


!/=/(«) 


The  sum  of  the  rectangles  of  which  DEQ'P  is  a  type,  is  approxi- 
mately equal  to  the  area  ABUV.  The  greater  n,  the  number  of 
rectangles,  i.e.,  the  smaller  Ax,  the  closer  will  the  sum  of  the 
rectangles  approximate  the  area  ABUV.     We  say  then  that 


_    lim 


1"^  2  DEQ'P, 


§65]  DIFFERENTIALS  103 

or 

x  =  b 


The  above  expression  (2)  represents  the  actual  area  and  not 
an  approximation  to  it,  as  can  be  shown  by  finding  the  greatest 
possible  error  corresponding  to  a  given  number  of  rectangles  and 
then  proving  that  this  error  approaches  zero  as  the  number 
of  rectangles  becomes  infinite.  Thus  it  is  easily  seen  that  the 
difference  between  the  true  area  A  and  the  sum  of  the  rec- 
tangles is  less  than  the  area  of  the  rectangle  RSTU.  The  altitude, 
f{b)  —  f{a),  of  this  rectangle  is  constant  while  the  length  of  the 
base,  Ax,  approaches  zero.  Hence  the  area  of  RSTU  approaches 
zero.  Therefore  the  limit  of  the  sum  of  the  rectangles  is  the 
area  sought. 

On  equating  the  two  expressions  for  A,  given  by  (1)  and  (2), 
we  have 

x  =  b 


where 


Fix)  =  ff(x)dx. 


This  equation  is  the  important  result  of  this  section.     It  gives  a 
means  of  calculating 


Ax= 

X  =a 


For,  to  calculate  this  limit  we  need  only  to  find  the  integral  of 
f{x)dx  and  take  the  difference  between  the  values  of  this  integral 
at  X  =  a  and  x  =  h.    The  result  of  this  section  will  be  restated 
and  emphasized  in  the  next  section. 
65.  Definite  Integral.    The  expression 

x  =  a 

which  was  introduced  in  the  preceding  section  is  of  such  great 


104  CALCULUS  [§65 

importance  that  it  is  given  a  name,  "the  definite  integral  of  f{x) 
between  the  limits  a  and  b,"  and  is  denoted  by  the  symbol 


i 


f{x)dx. 


Equation  (3),  §64,  gives  a  means  of  calculating  the  value  of  the 
definite  integral. 

The  function  F(x),  the  integral  oi  f{x)dx,  is  called  the  indefinite 
integral  of  j{x)dx  in  order  to  distinguish  it  from  the  definite  inte- 
gral which  is  defined  independently  of  it,  viz.,  as  the  limit  of  a 
certain  sum. 

We  have  then  the  following  definition  and  theorem: 

Definition.  Let  J{x)  be  a  continuous  junction  in  the  interval  from 
X  =  a  to  X  =  b,  and  let  this  interval  be  divided  into  n  equal  parts  of 
length  Ax  by  points  Xi,  Xz,  Xi,  .  .  .,  x„_i.  The  "definite  integral  of 
f{x)  between  the  limits  a  and  b"  is  the  limit  of  the  sum  of  the  products 
f{xi)  Ax  formed  for  all  of  the  points  Xq  =  a,  Xi,  xj,  .  .  . ,  x„_i,  as  the 
number  of  divisions  becomes  infinite. 

Theorem.  The  definite  integral  of  f(x)  between  the  limits  a  and 
b  is  calculated  by  finding  the  indefinite  integral,  F(x),  of  f{x)dx  and 
forming  the  difference  F{b)  —  F{a). 

The  symbol  for  the  definite  integral, 

fix)dx, 


r 


is  read  "the  integral  from  a  to  b  of  f{x)dx."     As  we  have  seen,  it 
means 

h 


Jf{x)dx  =  J2i  Zff{x)Ax. 
a  a 


Many  problems,  such  as  finding  the  work  done  by  a  variable 
force,  the  volume  of  a  solid,  the  coordinates  of  the  center  of 
gravity,  lead  to  definite  integrals.  But,  no  matter  how  a 
definite  integral  may  have  been  obtained  and  no  matter  what 
other  meaning  it  may  have,  it  can  always  be  regarded  as  repre- 
senting the  area  included  by  the  curve  y  =  f(x),  the  X-axis,  and 
the  ordinates  x  =  a  and  x  =  b,  provided  that  /(x)  is  a  function 


DIFFERENTIALS  105 

which  can  be  represented  by  a  continuous  curve.     This  fact,  that 

f{x)dx 


r 


can  be  regarded  as  representing  an  area,  enables  us  to  calculate 
its  value.  For  the  area  in  question  is  equal  to  F{b)  —  F{a), 
where  F{x)  is  the  indefinite  integral  of  f{x)dx.  Consequently  we 
have,  in  all  cases, 

f{x)dx  =  F{b)    -  F{a). 


r 


This  is  often  written 


X 


b 

j{x)dx  =  F{x) 


=  F{b)  -  Fia), 


to  show  how  the  result  is  to  be  calculated.     Thus 


I 


x^dx  =  -^ 


2^  _  P  _  7 
3        3  ~  3' 


Exercises 

Evaluate  the  following  definite  integrals: 

r  C'dx  C 

1.    I     (2x  +  Z)dx.  2.    I    ^-  3.    I    Va^  +  x"  xdx. 

Ji  Ji  Jo 

66.  Duhamel's  Theorem.  If  ai,  ocz,  as,  •  ■  •,  oin  are  n  in- 
finitesimals of  like  sign,  the  limit  of  whose  sum  is  finite  as  n  becomes 
infinite,  and  if  /3i,  jSz,  /Ss,  •  • ,  /3„  are  a  second  set  of  infinitesimals 
such  that 

lim   §J  _  1 
n=co  ^.  -  A. 

where  i  =  1,  2,  3,  •  •  •  ,  n,  then 

lim   V  „..  =   lim    V  p,. 

n=m    ^  n=co    ^  t^*' 

1  1 

Proof.   Let  —  =  1  +  €<. 

OCi 


106  CALCULUS  [§66 

Since 

lim   ^i  _  . 
n=co   „.        A. 

^^     €i  =  0. 

n=oo 

At  first  let  it  be  assumed  that  the  a's  are  positive.    Let  E  be  the 
numerical  value  of  the  largest  e,  i.e., 

E  >.  le<',         i  =  1,  2,  3,  •  •  • ,  n. 
Then,  since  /3i  =  «»  +  e.a,,         i  =  1,  2,  3,  •  •  • ,  n, 
ai  —  Eai  <  ^i  £  ai  +  Eai 
az  —  Ea-i  <.  ^1  ^  a2  +  Ea^ 

an   -   EUn    <^n<an-\-   E  Un- 

Adding,  we  get 

i  =  n  x=n  »'  =  » 

{l-E)%ai<^^i<{l-\-E)  X  cii. 
1=1  i  =  1  »■  =  1 

^^E  =  Q, 

n=oo 

n  i  =  n 

lim  ^  a<  =  lim  ^  /3i 
t  =  1  » =  1 

and  the  theorem  is  proved. 

If  the  a!s  are  negative,  it  will  be  necessary  to  change  the  proof 
just  given,  only  by  reversing  the  signs  of  inequality. 

Section  64  furnishes  an  illustration  of  this  theorem.  In  this 
example  the  limit  of  the  sum  of  the  infinitesimal  trapezoidal 
areas  DEQP  is  finite  as  n  becomes  infinite,  since  it  is  the  area 
sought. 

DEQ'P  <  DEQP  <  DEQP', 
(see  Fig.  46),  or 

yAx  <  DEQP  <  {y  +  Ay)  Ax, 
or 

DEQP        y  +  Ay_ 
^  ^    DEQ'P  ^       y 

This  shows  that  the  limit  of  the  ratio  of  the  trapezoidal  area  to 


Since 


n=< 
t  =  n 


§67]  DIFFERENTIALS  107 

the  area  of  the  corresponding  rectangle  is  1  as  n  becomes  infinite. 
Then  by  Duhamel's  Theorem, 

Um   2  DEQ'P  =    ^^^  2  DEQP  =  A, 

Since  we  are  able  to  replace  the  infinitesimals  DEQP  by  the 
infinitesimals  DEQ'P,  we  may  calculate  the  area  which  is  the  sum 
of  these  infinitesimals  by  means  of  the  definite  integral.  This 
is  a  characteristic  process  in  the  use  of  the  definite  integral. 
The  quantity  sought  is  subdivided  into  n  portions  which  are 
infinitesimals  as  n  becomes  infinite.  These  are  replaced  by  n 
other  infinitesimals  of  the  form  f{xi)  Ax.  The  limit  of  the  sum  of 
the  latter  infinitesimals  is  a  definite  integral. 

Since  the  limits  of  the  two  sums  are  equal  by  Duhamel's 
Theorem,  the  definite  integral  is  equal  to  the  quantity  sought. 

Illustrations  of  the  applications  of  Duhamel's  Theorem  to 
obtain  definite  integrals  representing  work,  force,  volume,  etc., 
follow. 

67.  Work  Done  by  a  Variable  Force.  In  §51  there  was  found 
the  work  done  by  a  variable  force, /(s),  in  producing  a  displacement 


Fig.  47. 

from  s  =  a  to  s  =  &.  We  shall  now  obtain  the  same  result  by 
building  up  the  definite  integral  which  represents  the  work. 
Divide  the  total  displacement  h  —  a,  Fig.  47,  into  n  equal  parts 
of  length  As.  The  force  acting  at  the  left  end  of  one  of  these 
parts  is  /(s),  while  that  acting  at  the  right  end  is  /(s  +  As). 
The  total  work  done  in  producing  the  displacement,  6  —  a,  is 
approximately 

8    =    6 

s  =  a 

The  actual  work  is  the  limit  of  this  sum  as  As  approaches  zero.^ 

'  This  step  can  be  justified  by  using  Duhamel's  Theorem.     Let  Au)  represent  the 
work  done  in  producing  the  displacement  As.     Then 


108  CALCULUS  [§67 

Illustration  1.     The  solution  of  the  problem  of  Illustration  1, 
§51,  is  expressed  by 

4 


I 


SOsds  =  15s2 


=  180. 

2 


Illustration  2.     The  solution  of  the  problem  of  Illustration  2, 
§51,  is  expressed  by 


w  = 


kmM    I     — -  =  —  kmM  —    =  —  kmM  \r I 

J„  s-  s  „  L6       a  J 


=  kmM\ 

Illustration  3.     In  solving  the  problem  of  Illustration  3,  §51, 
we  can  write 


_     lim 


«  =  o 


where  t)2  and  t^i  are  the  volumes  corresponding  to  s  =  a  and  s  —  b, 
respectively.     Since  pv''  =  C,  p  =  ^'  and 


,  C'dv  _      C       J 


C 


The  student  will  complete  the  numerical  work. 

n=  00 
But  /(s)As  <  Au)  <  /(s  +  As)A8, 

,  ^     Aw         /(s-fAs) 
°'  ^</(.)A«<-7(«) 

Then 

lim         Am 


1  _  A;    ^'"'~''  ~  '"'~''^- 


Hence  by  Duhatnel's  Theorem 
b 


As=0  /(s)A8 


So  2^-=   ii"o  2/WA.=/j'/(,M.. 

o  a  " 

s  =  b  c 

s  =  a  *f  "■ 


DIFFERENTIALS 


109 


Exercises 


1.  Set  up  and  evaluate  definite  integrals  representing  the  work 
sought  in  Exercises  1-5,  §51,  Chapter  VI. 

2.  Water  is  pumped  from  a  round  cistern  whose  median  section 
is  a  parabola.  The  cistern  has  a  diameter  of  8  feet  at  the  top  and 
it  is  16  feet  deep.  The  water  is  10  feet  deep.  Find  the  work  done 
in  pumping  the  water  from  the  cistern  if  the  discharge  of  the  pump 
is  3  feet  above  the  top  of  the  cistern  and  if  the  friction  in  the  pump 
and  the  friction  of  the  water  in  the  pipes  are  neglected. 

3.  Find  the  work  done  by  a  gas  in  expanding  in  accordance  with 
the  law  pv^-*  =  C  from  a  volume  of  10 
cubic  feet  to  one  of  12  cubic  feet,  if 
when  V  =  9  cubic  feet  p  =  100  pounds 
per  square  inch.  \B 

4.  Find  the  work  done  in  stretching 
a  spring  whose  original  length  was  15 
inches  from  a  length  of  16  inches  to  a 
length  of  18  inches  if  a  force  of  40 
pounds  is  required  to  stretch  it  to  a 
length  of  16  inches. 

6.  Find  the  work  done  in  compress- 
ing a  spring  of  original  length  5  inches 
to  a  length  of  3§  inches,  if  a  force  of 
900  pounds  is  required  to  compress  it 
to  a  length  of  4  inches. 

6.  The  force  due  to  friction  is  pro- 
portional to  the  component   of    force  Fig.  48. 
normal  to  the  surface  over  which  a  body 

is  being  moved.  Find  work  done  in  dragging  a  body  weighing  100 
pounds  from  the  base  to  the  top  of  a  slide  in  the  form  of  a  segment 
of  a  sphere,  Fig.  48,  if  the  distance  AB  =  200  feet  and  the  radius  of 
the  sphere  is  500  feet.  Express  the  result  in  terms  of  /*,  the  coeffi- 
cient of  friction. 

68.  Volume  of  a  Solid  of  Revolution.  The  area  bounded  by 
the  curve  y  =  f{x),  Fig.  46,  the  ordinates  x  =  a  and  x  =  b, 
and  the  X-axis,  is  revolved  about  the  X-axis.  Find  the  volume  of 
the  solid  generated. 

Divide  the  interval  AB  =  6  —  a  on  the  X-axis  into  n  equal  parts 
of  length  Ax  and  pass  planes  through  the  points  of  division 
perpendicular  to  the  X-axis.     These  planes  divide  the  volume  into 


H 


110  CALCULUS  [§68 

n  portions,  Ay.  A  typical  portion  can  be  regarded  as  generated 
by  revolving  DEQP,  Fig.  46,  about  the  base  DE  in  the  X-axis. 
Replace  the  volume  of  this  slice  by  that  of  the  cylinder  generated 
by  the  revolution  of  DEQ'P  about  the  Z-axis.  Its  volume  is 
TTj/'Ax.     The  total  volume  is  then 


or 


-r. 


Illustration.  Find  the  volume  between  the  planes  x  =  1  and 
X  =  3  of  the  solid  generated  by  revolving  the  curve  y  =  x^  -\-  x 
about  the  X-axis. 

/»3  /»3 

F  =  TT  I       {X^  +  Xydx  =  TT  \       {x*  +  2x3  _|_  x^)(lX 

=  X[U'  -I-  \X*  -I-  W]\    =  TTXV^X^  +  hx  +  ii; 

=  27ir(|  -1-3-^1)-  x(i  +  i  4-  i)  =  -^ tPr. 

Exercises 

1.  Find  the  volume  between  the  planes  x  =  0  and  x  =  3  of  the  solid 
generated  by  revolving  the  parabola  y^  =  6x  about  the  X-axis. 

2.  Find  the  volume  of  a  sphere  of  radius  r. 

3.  Find  the  volume  of  the  ellipsoid  of  revolution  generated  by 
revolving  the  ellipse 

16  "^  9 

about  the  X-axis;  about  the  K-axis. 

4.  Find.the  volume  between  the  planes  x  =  0  and  x  =  4  of  the  solid 
generated  by  revolving  y^  =  x^  about  the  X-axis. 

6.  Find  the  volume  of  the  solid  generated  by  revolving  x'  +  y^  =  a' 
about  the  X-axis. 

6.  Find  the  volume  generated  by  revolving  y^  =  2ax  —  x*  about 
the  X-axis. 

7.  Find  the  volume  generated  by  revolving  the  oval  of 
y*  =  x(x  —  l)(x  —  2)  about  the  X-axis. 


§69] 


DIFFERENTIALS 


111 


69.  Length    of   Arc:   Rectangular   Coordinates.     In    §63    the 

length  of  arc  of  a  curve  was  found  by  integrating  its  differential. 
We  shall  now  express  the  length  of  arc  by  means  of  a  definite 
integral. 

To  find  the  length  of  arc  APQB,  Fig.  49,  divide  CH  into  n 
equal  parts  of  length  Ax  each.  At  the  points  of  division  erect 
ordinates  dividing  the  arc  AB  into  n  parts  of  which  PQ  is  one. 
The  length  of  arc  AB  is  defined  by 


g  ^  lim  V  ^ 

n=oo  ^^ 


3 

P   Q 

A 

/ 

y 

^ 

AC 

\ 

B 

\ 

O 

C               D   E 

■«-o-*j 

H- 

Fig.  49. 
where  Ac  is  the  length  of  the  chord  PQ.     Then 
s  =   ^^^  y,V(Ax)^+(Ayy 


Since 


=  1, 


it  follows  by  Duhamel's  Theorem  that 

b 

lim 
s 


=:iT.x^^uM'^ 


112  CALCULUS  [§70 

Hence, 

«  =    I     \/l  +  l^)  dx. 


r>/^^ 


Exercises 

l.^Find  the  length  of  the  curve  y  =  x^  between  the  points  (0,  0) 

and'(l,  1). 

z         1         1 

2.  Find  the  entire  length  ofx*  +y*  =  a^- 

3.  Find  the  entire  length  of  x^  +  y^  =  a^. 

4.  Find  the  length  of  y^  =  4:X^  between  the  points  (0,  0)  and  (4,  16). 
70.  Area  of  a  Surface  of  Revolution.     The  portion  AB,  Fig.  49, 

of  the  curve  y  =  f{x),  between  the  ordinates  x  =  a  and  x  =  b, 
is  revolved  about  the  X-axis.  Find  the  area,  S,  of  the  surface 
generated. 

Pass  planes  as  in  §69  perpendicular  to  the  X-axis  through  the 
equidistant  points  of  division  of  the  interval  CH  =  h  —  a. 
Denote  the  convex  surface  of  the  frustum  of  the  cone  generated  by 
the  revolution  of  DEPQ  by  AF.  The  area,  S,  of  the  surface  of  re- 
volution will  be  defined  as  the  limit  of  the  sum  of  the  convex 
surfaces,  AF,  of  these  frusta  as  n  becomes  infinite,  i.e.,  as  Ax 
approaches  zero.     Then, 


„        lim  V  A  p        lim  V  o    ^  +  ('^  +  ^V^ 


Ac 


Ay 
By  Duhamel's  Theorem  we  can   replace   y  +  -„"    by  y,  since 


lira  2/ +  |Aj/ 
Ax-o V, =  1-     Hence, 


Since 


lim , — , 

\x=0      I T —  =  1> 

'dy\i 


§71]  DIFFERENTIALS  113 


Therefore 


a 

yds, 

x  —  a 


5  =  2x  I    v\il  +  [~\''dx 


=  27r 

where  ds  is  the  differential  of  the  length  of  arc.  The  latter  form 
is  easily  remembered  since  2Tryds  is  the  area  of  the  strip  of  surface 
generated  by  revolving  ds,  the  differential  of  arc,  about  the  X-axis 
at  a  distance  y  from  it.  If  it  is  more  convenient  to  integrate  with 
respect  to  y,  ds  can  be  replaced  by 


v^d)^"- 


and  the  limits  are  the  values  of  y  corresponding  to  x  =  a  and  x  =  b. 
Thus 

Wl  +  U)    dy  =  2ir\  yds. 

Exercises 

1.  Find  the  surface  between  the  planes  x  =  0  and  x  =  5  of  the 
paraboloid  of  revolution  obtained  by  revolving  t/'*  =  4x  about  the 
X-axis. 

2.  Find  the  surface  of  the  sphere  generated  by  revolving  x''  +  2/^  =^* 
about  the  X-axis. 

3.  Find  the  surface  of  the  right  circular  cone  whose  altitude  is  10 
feet  and  the  radius  of  whose  base  is  5  feet. 

4.  Find  the  surface  of  the  solid  generated   by  the  revolution  of 

2  2  2 

a^    +  2/'  =  a*  about  the  X-axis. 

71.  Element  of  Integration.  The  first  step  in  setting  up  a 
definite  integral  is  to  break  up  the  area,  volume,  work,  length,  or 
whatever  it  is  desired  to  calculate,  into  convenient  parts  which 
are  infinitesimals  as  their  number  approaches  infinity.  These 
parts  are  then  replaced  by  other  infinitesimals  of  the  typical 

8 


114 


CALCULUS 


[§72 


form  fix)dx,  which  must  be  so  chosen  that  the  limit  of  the  ratio 
of  each  infinitesimal  of  the  second  set  to  the  corresponding 
infinitesimal  of  the  first  set  is  one.  fix)dx  is  called  the  "element" 
of  the  integral  or  of  the  quantity  which  the  integral  represents. 
Thus  the  element  of  volume  is  wy^dx,  that  of  area  is  ydx,  that  of 
work  is  Fdx. 

If  the  magnitude  which  it  is  desired  to  calculate  is  broken  up  into 
suitable  parts,  the  expressions  for  the  elements  can  be  written 
down  at  once.  The  best  way  of  retaining  in  mind  the  formulas 
of  §§68,  69,  and  70  is  to  understand  thoroughly  how  the  elements 
are  chosen.  The  process  of  writing  down  the  element  of  integra- 
tion at  once  becomes  almost  an  intuitive  one. 

72.  Water  Pressure.  The  pressure  at  any  given  point  in  a 
liquid  at  rest  is  equal  in  all  directions.    The  pressure  per  unit 

area  at  a  given  depth  is  equal  to  the 
pressure  on  a  horizontal  surface  of 
unit  area  at  that  depth,  i.e.,  to  the 
weight  of  the  column  of  liquid  sup- 
ported by  this  surface.  This  weight 
is  proportional  to  the  depth.  Hence 
the  pressure  at  a  depth  x  below  the 
surface  of  the  liquid  is  given  by  the 
formula  p  =  kx.  If  the  liquid  is 
water  and  the  depth  x  is  expressed 
in  feet,  k  =  62.5  pounds  per  cubic 
foot. 
The  method  to  be  used  in  finding  the  water  pressure  on  any 
vertical  surface  is  illustrated  in  the  solution  of  the  following 
problems: 

1.  Find  the  pressure  on  one  side  of  a  gate  in  the  shape  of  an 
isosceles  triangle  whose  base  is  6  feet  and  whose  altitude  is  5  feet, 
if  it  is  immersed  vertically  in  water  with  its  vertex  down  and  its 
base  4  feet  below  the  surface  of  the  water. 

Take  the  origin  at  the  vertex  of  the  triangle,  the  axis  of  x  vertical, 
and  the  axis  of  y  horizontal,  as  in  Fig.  50.  The  altitude  is  sup- 
posed to  be  divided  into  n  equal  parts  and  through  the  points  of 
division  horizontal  lines  are  supposed  to  be  drawn  dividing  the 
surface  into  strips.     The  trapezoid  KHMN  =  AA  is  a  typical 


§72]  DIFFERENTIALS  115 

strip.  Denote  the  pressure  on  this  strip  by  AP.  The  abscissa 
of  the  lower  edge  of  the  strip  is  x  and  the  pressure  at  this  lower 
edge  is  A;(9  —  x).    Then  the  total  pressure  is 

P  =  n^^Xm-x)LA.  (1) 

In  accordance  with  Duhamel's  Theorem  we  can  replace  AA 
by  2i/Ax. 

x  =  0 


or 


Since 


=  2A;  I    (9  - 


x)ydx.  (3) 


V  = 


3x 
V 


•5 

P  =  ~  \    (,9-x)xdx 


_6kr 

5  Jo 

=  g-  "2-  -  3^      =  5312.5  pounds.  (4) 

In  general,  if  u  denotes  the  depth  below  the  surface  of  the  liquid 
and  z  denotes  the  width,  at  the  depth  u,  of  the  vertical  surface 
on  which  the  pressure  is  to  be  computed, 

P  =  k  \    uzdu,  (5) 

where  a  and  b  are  the  depths  of  the  highest  and  lowest  points, 
respectively,  of  the  surface.     For, 

p  =    lim  X\kuAA  =   lin^     X  kuzAu  =  k  \    uz  du. 

n^a,  ^^  Au=0   •^T  I 

2.  Find  the  total  pressure  on  a  vertical  semi-elliptical  gate 
whose  major  axis  lies  in  the  surface  of  the  water,  given  that  the 
semi-axes  of  the  ellipse  are  8  feet  and  6  feet.     Take  the  origin  at 


116 


CALCULUS 


[§73 


the  center,  the  axis  of  x  horizontal  and  the  axis  of  y  positive  down- 
ward.    The  element  of  pressure  is 


and  the  total  pressure  is 


2kyx  dy 


^  =  4 


yxdy. 


X  is  expressed  in  terms  of  y  by  means  of  the  equation  of  the 
ellipse, 


Then 


—  -u  ?^  =  1 
64  "^  36 


P  =  2ki  I  yVSQ  -  y^  dy. 


Exercises 

1.  Find  the  pressure  on  the  vertical  parabolic  gate,  Fig.  51:  (o) 
if  the  edge  AB  lies  in  the  surface  of  the  water;  (b)  if  the  edge  AB  lies 
5  feet  below  the  surface. 


2.  Find  the  pressure  on  a  vertical  semicircular  gate  whose  diameter, 
10  feet  long,  lies  in  the  surface  of  the  water. 

73.  Arithmetic  Mean.  The  arithmetic  mean,  ^,  of  a  series  of 
n  numbers,  Oi,  02,  aa,  •    •    •  ,  a„,  is  defined  by  the  equation 


or 


nA  =  ai  +  a2  +  as  +  •    •    •  +  a„, 
ai  +  a2  +  as  +  •    •    •  +  a„ 


A  = 


That  is,  A  is  such  a  number  that  if  each  number  in  the  sum 


§741  DIFFERENTIALS  117 

Oi  +  c[2  +  cfs  +  *  •  '  -\-  ttn  be  replaced  by  it,  this  sum  is 
unaltered. 

74.  Mean  Value  of  a  Function.  We  can  extend  the  idea  in- 
volved in  the  arithmetic  mean  to  other  problems. 

Illustration  1.  Suppose  a  body  moves  with  uniform  velocity 
a  distance  of  1  foot  during  the  first  second,  a  distance  of  2  feet 
during  the  second  second,  a  distance  of  3  feet  during  the  third 
second,  and  so  on  for  10  seconds.  At  the  end  of  10  seconds  the 
body  would  have  moved  1  +  2  +  3+  •  •  -+10  =  55  feet. 
The  mean,  or  average,  velocity  of  the  body  is  the  constant  velocity 
with  which  the  body  would  describe  this  distance  in  the  same 
time.     It  is  equal  to  5.5  feet  per  second. 

If  the  velocity  of  the  body  instead  of  changing  abruptly  as 
indicated  above  were  changing  continuously  in  accordance  with 
the  law  V  =  t,  the  total  distance  s  traversed  in  10  seconds  would  be 

•10 

tdt  =  50. 


J "10  /*1 

vdt  =    j 
0  Jo 


From  this  equation 


The  mean  velocity,  V,  the  constant  velocity  which  a  body  must 
have  in  order  to  traverse  the  same  distance  in  the  same  length 
of  time,  is  50  -i-  10  =  5  feet  per  second.  This  can  be  expressed 
by  the  formula 

J "10  /»10 

Vdt  =    I     vdt. 
0  Jo 

vdt 

10 

In  general,  if  i;  =  j{t),  the  mean  velocity,  V,  of  the  body  in 
the  interval  of  time  between  t  =  a  and  f  =  6  is  expressed  by  the 
equation 

=    { At)dt, 


or,  since  F  is  a  constant, 

S{t)dt 
V  = 


Vdt=    I 

a  Ja 

Ja 


b  —  a 


118  CALCULUS  [§74 

V  is  the  constant  velocity,  which  replacing  the  variable  velocity, 
"  =  /(O,  at  every  instant  in  the  interval  between  t  =  a  and  t  =  b, 

gives  the  same  distance  traversed,  i.e.,  leaves  the  value  of  the 

nb 
integral,    I    v  dt,  unchanged. 


Illustration  2.  Consider  the  work  done  by  a  variable  force  / 
acting  in  a  straight  hne,  the  X-axis,  and  producing  a  displacement 
from  a;  =  a  to  X  =  6.  If  the  law  of  the  force  is  f  =  (f){x),  the 
mean  force  F  in  the  interval  from  a;  =  a  to  x  =  6,  or  the  constant 
force  which  would  do  the  same  work  while  producing  the  dis- 
placement 6  —  o,  is  given  by  the  equation 


nb  f*b 

I  Fdx  =  I   0(x)dx, 

f. 


or 

(l){x)dx 

F  = 

b  —  a 

F  is  a  constant  such  that  if,  in  the  integral  I  4>{x)dx,  the  func- 
tion ^(x)  be  replaced  by  it,  the  value  of  the  integral  remains 
unchanged. 

Illustration  3.  Let  a  unit  of  mass  be  situated  at  each  of  the 
points  on  the  X-axis  whose  abscissas  are  Xi,  X2,  X3,  •  •  •,  x„. 
The  X-axis  is  taken  horizontal  and  the  masses  are  acted  upon 
by  gravity.  We  shall  find  the  distance,  x,  from  the  origin  at 
which  the  n  masses  must  be  concentrated  in  order  that  the  sum  of 
the  moment  about  the  origin  of  the  forces  acting  on  the  masses 
shall  be  unchanged. 

Clearly  x  must  satisfy  the  equation 

gnx  =  g{xi  +  X2  +  X3  -{-  •    •    •  +  x„), 
or 

Xi  +  X24-X3+  •   •   .  +x„ 

X  =  • 

n 

If  there  are  mi,  mj,  ms,  •  «  •  ,  7n„  units  of  mass  concentrated 
at  Xi,  X2,  Xs,  •  •  •  ,  x„,  respectively,  the  mean  moment  arm,  x, 
the  distance  from  the  origin  at  which  the  masses  must  be  con- 


§74]  DIFFERENTIALS  119 

centrated  in  order  that  the  sum  of  the  moments  about  the  origin 
of  the  forces  acting  on  the  masses  shall  be  unchanged,  is  given  by 
the  equation 

(mi  +  m2  +  •    •    •  m„)x  =  niiXi  +  W2X2  +  •    •    •  +  m„x„, 
or 

-  2  THiXi 


2wi 


-,  (i=  1,2,3,.    .    .,n).  (1) 


rr  is  a  constant  such  that  if  in  the  sum  2  m,x,  each  of  the  num- 
bers Xi,  X2,  •    •    •  ,  x„  be  replaced  by  x  this  sum  is  not  changed. 

Now  let  there  be  a  continuous  distribution  of  matter  along  the 
X-axis  from  x  =  a  to  x  =  b.  Divide  the  interval  b  —  a  into  n 
segments  each  of  length  Ax.  An  expression  for  the  approximate 
sum  of  the  moments  about  the  origin  of  the  forces  acting  on  the 
mass  is  2  gxAmt,  where  Am,  is  the  mass  of  the  segment  AZf. 
An  expression  for  the  approximate  force  is  2  ^Am,-.     Hence  an 

expression  for  the  approximate  x  is   „    . — -*•     It  is  readily  seen 

that  as  Ax  approaches  zero,  the  numerator  approaches  the  total 
moment  and  the  denominator  approaches  the  total  mass.     Hence 


lim  ^   ^  T^ 

Am=o  ZigAm  I   dm 


(2) 


X  is  a  constant  such  that  if  in  the  integral,  I   x  dm,  x  is  replaced 


,  I    xdm, 


by  X,  the  value  of  the  integral  is  unchanged. 

For  example,  if  the  density  is  proportional  to  x^,  i.e.,  is  equal  to 
kx^,  the  element  of  mass,  dm,  is  kxHx,  and  we  have 


J'*b  nb 

xkxHx  I   x^dx 

_        a  _     Ja 


•6  Ch 

Mx 

3  b*  -  a\ 

»b  7'6  4  fe»  —  a' 


kx^dx  I    x^dx 

a  Ja 

The  mean  value,  M,  cf  the  function  fix)   with  respect  to  the 


120  CALCULUS  [§74 

magnitvde  u,  which  is  a  function  of  x,  is  defined  by  the  equation 

(3) 

where  M  is  a  constant,  or 


Jf*x  =  b  r*x  =  b 

Mdu  =    I  f{x)du, 
x  =  o  Jx  =  a 

J'*x  =  b 
f{x)du 
x  =  a 


M  = 


'1  =  6 

du 


J^di 
x  =  a 


(4) 


M  is  a  constant  such  that  if  the  function  f{x)  is  replaced  by  it  in 

Jr*x  =  b 
f{x)du,  the  valv£  of  the  integral  is  not  changed. 
a;  =  o  _ 

In  (2),  X  is  the  mean  value  of  x  with  respect  to  the  magnitude, 
m. 

A  particular  case  of  (4)  is  that  in  which  u  =  x.  Then  (4) 
becomes 

M  =  ,— ^^    I    f (x)  dx.  (5) 


1       p 


Illustrations  1  and  2  are  cases  of 
this  type. 

When  w  =  a;,  as  in  equation 
(5),  M  can  be  interpreted  as 
the  altitude,  AC,  of  a  rectangle 
wth  base  AB  =  b  —  a,  Fig. 
52,  whose  area  is  equal  to  the 
area  bounded  by  the  curve  y  = 
■X  f{x),  the  A'-axis,  and  the  ordi- 
nates  x  =  a  and  x  =  b.  From 
this  standpoint  M  is  called  the 
mean  ordinate  of  the  curve  y  = 

f{x)  in  the  interval  from  x  =  a  to  x  =  b. 
Illustration  4.     Find  the  mean  ordinate  of  the   curve  y  =  x' 

between  the  ordinates  a;  =  0  and  x  =  2. 


Fig.  52. 


M  =  ^-^   I    f{x)dx  =  §  I   x^dx  =  ix3 


4. 


§74]  DIFFERENTIALS  121 

Illustration  5.  Find  the  mean  with  respect  to  w  of  x  between 
the  limits  u  =  \  and  tt  =  9,  if  m  =  x^. 

J'*u  =»  9  (*u  =  9 

Mdu  =    I   xdu, 
u  = 1  Ju  = 1 

J  3  /»3 

2xdx=    I   2x»dla:, 

8M  =  V, 
M  =  V. 

Exercises 

Find  the  mean  ordinates  for  the  following  curves: 

1.  y  =  X*  between  x  =  0  and  x  =  3. 

2.  2/  =  x^  between  x  =  2  and  x  =  4. 

3.  2/  =  3x3  between  x  =  0  and  x  =  2. 

4.  y  =  3x3  between  x  =  1  and  x  =  3. 

5.  J/  =  x*  between  x  =  0  and  x  =  1. 

6.  Find  the  radius  of  the  right  circular  cylinder  of  altitude  3  whose 
volume  is  equal  to  the  volume  between  the  planes  x  =  2  and  x  =  5  of 
the  solid  generated  by  revolving  7/  =  x  +  x''  about  the  X-axis. 

7.  Find  the  radius  of  the  right  circular  cylinder  of  altitude  h  —  a 
whose  volume  is  equal  to  the  volume  between  the  planes  x  =  a  and 
X  =  fe  of  the  solid  generated  by  revolving  y  =  /(x)  about  the  X-axis. 

8.  The  density  cf  a  thin  straight  rod  10  inches  long  and  of  uniform 
cross  section  is  proportional  to  the  distance  from  one  end.  Find  the 
mean  density  of  the  rod. 

9.  Find  the  mean  velocity  of  a  freely  falling  body  between  the 
time  ^  =  1  second  and  t  =  3  seconds. 

10.  The  density  of  a  rod  is  given  by  p  =  3x2,  'v^here  x  is  the  distance 
from  one  end.     Find  the  mean  density  if  the  rod  is  10  inches  long. 

11.  Find  the  mean  moment  arm  in  the  case  of  the  rods  of  Exercises 
8  and  10,  about  a  horizontal  axis  through  the  end  of  the  rod  (x  =  0). 
The  rods  are  horizontal,  and  perpendicular  to  the  axis  about  which 
moments  are  taken.  The  rods  are  supposed  to  be  acted  upon  by 
forces  due  to  gravity  alone. 

12.  Find  the  mean  ordinate  of  a  semicircle,  the  ends  of  which  are 
upon  the  X-axis. 


CHAPTER  VIII 
CIRCULAR   FUNCTIONS.     INVERSE    CIRCULAR   FUNCTIONS 

Up  to  this  point  only  functions  have  been  discussed  which  are 
simple  algebraic  combinations  of  powers  of  the  dependent  variable. 
Many  interesting  applications  of  the  calculus  to  the  study  of 
these  functions  have  been  given.  We  shall  now  take  up  the 
study  of  the  appUcation  of  the  methods  of  the  calculus  to  another 
very  important  class  of  functions,  the  circular  functions.  It  is 
apparent  that  the  principles  developed  in  the  preceding  chapters 
are  equally  applicable  to  the  circular  functions  and  to  the 
algebraic  functions. 

As  the  student  has  already  learned,  the  circular  functions  occur 
very  frequently  in  the  study  of  the  physical  sciences  and  their 
applications,  because  by  means  of  them  periodic  phenomena  can 
be  studied. 

76.  Derivative  of  sin  u. 
Let 

y  =  sin  u. 
y  -\-  Ay  =  sin  (u  +  Am), 

Ay  =  sin  {u  +  Aw)  —  sin  u 

—  sin  u  cos  Au  +  cos  u  sin  Au  —  sin  u, 
Ay  _  cos  w  sin  Am       sinM(l  —  cosAm) 
Aw  Am  Am 

Then 

lim  A^  _  lim  sin  Am        .       lim  1  -  cos  Am 

A«=o  Am  ~  ^°^  "^"=0  Am         ®^°  ^^«-o       Am       * 

Hence  by  §56  and  §58 

dy 

^  =  cosw..  (1) 

Whence 

dy  du 

di  =  «°«^dx"  (2) 

122 


§75] 


CIRCULAR  FUNCTIONS 


123 


The  corresponding  formula  for  dy  is 
dy  =  cos  udu. 
It  has  thus  been  shown  that 


d(smu)  du 

— J =  cos  u  j- 

dx  dx 


(3) 
(4) 


and 


d(sinu)  =  cos  udu. 

Well  known  properties  of  the  function  y  =  sin  u  can  be  verified 
by  formula  (1).     Thus  sin  u  is  an  increasing  function  between 

tt  =  0  and  u  =  t),  and  between  u  —  -^  and  u  =  27r,  and  decreas- 
ing  between  w  =  ^  and  w  =  -p"    The  same  facts  are  shown  by 

re  I    I    I    p^  I    I    I    I    I    I    I    I    I    I    I 

G_u 


Fig.  53. 


the  derivative,  cos  u,  which  is  positive  between  w  =  0  and  «  =  o' 


Stt 


and  between  u  =  -^  and  u  =  27r,  and  negative  between  u  =  ^ 


and  u    = 


Stt 


Further,    sin   u   has   maximum   and   minimum 

values  for  u  =  ^  and  u  =  -^»   respectively.     The   same  facts 

are  shown  by  the  derivative,  cos  u,  which  becomes  zero  at  these 

points  and  changes  sign  at  ^  from  plus  to  minus,  and  at  -^  from 

minus  to  plus. 

The  slope  of  the  sine  curve  is  approximately  the  slope  of  the 
diagonal  PQ  of  a  rectangle  in  Fig.  53.  The  greater  the  number  of 
equal  parts  into  which  the  circumference  of  the  circle  is  divided 
and  hence  the  smaller  the  subdivisions  of  the  arc,  the  closer  do  the 
slopes  of  these  diagonals  approach  the  slopes  of  the  tangents. 


124  CALCULUS  [§76 

76.  Derivatives  of  cos  u,    tan  u,    cot  u,    sec  u,    esc  u.     The 

derivatives  of  the  remaining  circular  functions  can  be  obtained 
from  that  of  the  sine. 
Let  y  =  cos  u.    Then 


y  =  sin 


and 


dy              /x          X^(I-^) 
^  =  cos^2-i.j-   ^^^ 

du\ 


=  cos  (I  -  u)  (-  ^) 
du 


=  —  sm  «  -r' 
dx 

Hence 

d(cos  u)             .       du 

— ^ =  —  sm  u  :i— 

dz                       dz 

and 

d(cos  u)  =  —  sin  u  du. 

By  writing 

sin  u 

tan  u  =  » 

cos  u 

cos  u 

cot  u  =  --. » 

sin  u 

and 


sec  u 


1 

CSC  u  = 


(1) 


sin  u 
the  student  will  show  that 

d(tan  u)  du 

— ^ =  sec^'u^  »  or    d(tan  u)  =  sec^u  du  (2) 

d(C0t  U)                            du                       Ar     4.     \                      2       A  f}\ 

— J =  —  csc'^u  ,   »  or   d(cotu)  =  —  csc^  u  du  (3) 

d(sec  u)                *         du            ^r         \               *.         a  /^^ 

— -V =  sec  u  tan  u  ,  '  or   d(sec  u)  =  sec  u  tan  u  du  (4) 

d(cscu)                      .     du           ,.         .                        ,      ,  ._. 

- — J =  —  CSC  u  cot  u  J  »  or   d(csc  u)  =  —  csc  u  cot  u  du  (5) 


§76]  CIRCULAR  FUNCTIONS  125 

Illustration  1.     Find    the    first    and    second    derivatives    of 
3  sin  (2a;  -  5). 

rf[3sin  (2a; -5)]  _      d[sm  (2x  -  5)] 
dx  dx 

d(2x  -  5) 


dx 


=  3  cos  (2a;  -  5) 

=  6  cos  (2a;  —  5). 
Differentiating  again, 

d'[3sin  (2a; -5)]  ^     rf[cos(2x-5)] 
dx^  dx 

=  —6  sin  (2x  —  5) ^ 

=  -12sin(2x  -  5). 

dv 
Illustration  2.     li  y  =  sin  2a;  cos  x,  find  j—     Since  sin  2x  cos  x 

is  the  product  of  two  functions,  apply  formula  (1)  §40. 

dy        •     „  /       .       ^  dx   .    ,         X  /       ^  X  d2x 
-T-  =  sin  2a;(  —  sin  x)-i — |-  (cos  x)  (cos  2x)  -r- 

=  2  cos  x  cos  2a;  —  sin  x  sin  2x. 

dv  d^v 

Illustration  3.     If  ?/  =  3  sin  x  +  4  cos  x,  find  --r-  and  -r-^' 

-p  =  3  cos  X  —  4  sin  X  (6) 


From  (6) 


d^y 

J— 2  =  —(3  sin  X  +  4  cos  x)  =  —y.  (7) 

-r-  =  4  cos  X  (f  —  tan  x). 


TT 

When  0  <  x  <  ^,  cos  x  is  positive.    The  second  factor,  f  —  tan  x, 

is  positive  when  x  <  tan~^  (^),  and  negative  when  x  >  tan"^  (f ). 
Thus,  when  x  is  in  the  first  quadrant  the  function  has  a  maximum 
value  coriiesponding  to  x  =  tan~i  (|). 

When  r,  <  a;  <  TT,  -T-  is  negative. 

3x 
When  TT  <  X  <  -^,  cos  x  is  negative,  and  J  —  tan  x  is  negative 


126 


CALCULUS 


[§76 


when  X  <  tan~^  (f),  and  positive  when  x  >  tan-'  (f).  Thus 
when  X  is  in  the  third  quadrant  the  function  has  a  minimum 
value  corresponding  to  the  value  x  =  tan~^  (j). 

When  -^  <  x<  lir,  -3-  is  positive. 

The  same  facts  can  be  seen  directly  from  the  function,  for  it 
can  be  put  in  the  form 

y  =  5(5  cos  X  +  t  sin  x). 


Let  cos  a  =  i  and  sin  a 


Then 


or 


y  =  5(cos  x  cos  a  +  sin  x  sin  a), 
y  =  5  cos  (x  —  a). 


In  polar  coordinates  this  represents  a  circle  passing  through  the 

origin,  with  a  diameter  of  5.     (See  Fig.  54.)     x  is  the  vectorial 

angle  and  y  the  radius  vector.     The 

diameter  OB  makes  an  angle  a  with 

the  polar  axis.     As  x  varies  from  0 

to  IT  the  circle  is  described,  and  as  x 

varies  from  tt  to  27r,  y  is  negative 

and  the  circle  is  described  a  second 

time. 

Hence  y  has  a  maximum  value  5 

when  X  is  equal  to  a,  and  a  mini- 

FiG.  54.  mum  value  —5  when  x  is  equal  to 

a  -\-  IT. 

dy  d^y 

Illustration  4.     li  y  =  tan'  3x  =  (tan  ZxY,  find  -r-  and  -j-^* 

The  function  is  of  the  form  y  =  w.     Hence 
g  =  3(tan3x)^^^^^ 


=  3  tan^  3x  sec^  3x 


d3x 
dx 


=  9  tan^  Zx  sec^  3x. 


d^ 
dx* 


9    tan*  3x 


c?(sec*  3x) 
dx 


+  sec*  3x 


(i(tan*  3x)- 
dx 


§76]  CIRCULAR  FUNCTIONS  127 


=  18  r 


tan2  3x-2  sec  3a;    ^    ,       '  +  sec^  3x-2  tan  3x  -  ■ 


dx  dx 

dSx 


tan^  3a;  sec  3a;  sec  3a;  tan  3a;    , 

dx 

dSx^ 
+  sec'^  3x  tan  3a;  sec^  3x  -i — 

=  54  (tan'  3x  sec^  3a;  +  tan  3x  sec*  3a;) 
=  54  tan  3a;  sec''  3x(tan2  Sx  +  sec'  32). 

Illustration  5.     If  -r-  =  cos  x,  find  y. 
dy 

-T-  =  COS  x. 

dx 

y  =  sinx  -\-  C. 

dv 
Illustration  6.     If  -p  =  cos  3a;,  find  y. 

dy      .r        „    d3a;"| 
^=i[cos3x^J. 

The  expression  within  the  bracket  is  the  derivative  of  sin  3x, 

hence, 

2/  =  ^  sin  3x  +  C. 

dv 
Illustration  7.     If  ",^  =  sin  3a;,  find  y. 


Hence 


=  -i[-sin3a;^]. 


?/  =  —  ^  cos  3a;  +  C. 
Illustration  8.     If  -j-  =  sec^  2x,  find  y. 

1/  =  5  tan  2a;  +  C. 

dv 
Illustration  9.     If  -^-  =  sec  5x  tan  5x,  find  y. 


Hence 


=  I   sec  5x  tan  5x  -r—   . 


128  CALCULUS  [§76 

Hence 

y  =  6  sec  5a;  +  C. 

Illustration  10.     li  dy  =  cos  3x  dx,  find  y. 

y  =   \  cos  Zxdx 
=  \  \  cos  3x  rf(3x) 
=  \  sin  3x  +  C'. 
Illustration  11.     li  dy  =  sin'^  2x  cos  2x  dx,  find  ?/. 

y  =  j  (sin  2x)^  cos  2x  dx 
=  i  5/3 (sin  2x)2  cos  2x  rf(2x) 

=  i  J3(sin  2xyd{sm  2x). 

Hence 

2/  =  Ksin2x)«  +  C. 

Illustration  12.     If  d?/  =  tan^  5x  sec^  5x  dx,  find  y. 
y  =  j  tan^  5x  sec^  5x  dx 
=  i  i J  4  (tan  5x)^  sec''  5x  d(5x) 

=  "sHr  j  4  (tan  5x)'  d(tan  5x). 

Hence 

y  =  ■2V(tan  Sx)^  +  C. 
Illustration  13. 
J  sin  5x  cos  3x  dx  =  J  Msin  (5x  +  3x)  +  sin  (5x  —  3x)]dx 

=  5  I  sin  8x  dx  +  ^  J  sin  2x  dx 

=  —  iV  cos  8x  —  J  cos  2x  +  C. 
Illustration  14. 
j  cos  7x  sin  3x  dx  =  J  |[sin  (3x  +  7x)  +  sin  (3x  —  7x)]dx 

=  I J  sin  lOx  dx  —  5  J  sin  4x  dx 
=  —  2V  cos  lOx  +  I  cos  4x  +  C. 


§76]  CIRCULAR  FUNCTIONS  129 

Illustration  15. 
J  cos  4a;  cos  7x  dx  =  J  |  [cos  (7x  +  4a;)  +  cos  (7x  —  4a;)]  dx 

=  5  J  cos  llx  dx  +  hi  cos  3a;  dx  +  C 

=  -2V  sin  11a;  +  6  sin  3a;  +  C. 
Illustration  16. 

J  sin  4x  sin  2x  dx  =  —  ^  J  [cos  (4a;  +  2x)  —  cos  (4a;  —  2x)]  dx 

—  ~  2]  cos  6x  dx  -{-  h  j  cos  2a;  rfx 
=  —  iV  sin  6x  +  J  sin  2x  +  C. 

Exercises 

In  Exercises  1  to  10,  verify  the  differentiation. 

1.  y  =  sin  ox,  ^  =  5  cos  ox,         -3-^  =  —  2oy. 

dy  „    .    „       d^y 

2.  y  =  cos  3a;,  dx  ^  ~     ®'"      '    dx^  "  ~  ^^* 

3.  y  =  tan  2x,  'd    ^ '^  ^^^^  ^^' 

j-^  =  8  sec''  2x  tan  2x. 

dw 

4.  2/  =  sin  X  cos  2x,         -r-  =  cos  2x  cos  x  —  2  sin  2x  sm  x 

.     3x  -  2  dy        ,  3x  -  2 

6.  7/  =  sin  — ^ —  =  I  cos 


dx        '  5      ' 

dx-^  =  -  ^'  y- 

6.  2/  =  tan'  5x,  dy  =  15  tan^  5x  sec*  5x  dx. 

7.  y  =  sec^  3x,  dy  =  12  sec*  3x  tan  3x  dx. 

0     [y  ~  °(1  ~  cos  0),    dy  =  a  sin  0  d9. 

Lx  =  a{d  —  sin  0),     dx  =  a(l  —  cos  O)d0. 

.    /2wt        \    dy       2aw  /27rt        \ 

9.  y  =  a  sin  I  ^^ ^)  '  di  ^  ^  ^°^  XT'  ~  V  ' 

10.  y  =  X  sin  x,  dy  =  (x  cos  x  +  sin  x)  dx. 

11.  From  the  results  of  Exercise  8,  show  that  -r-  =  cot  s- 

'  dx  2 

0 


130  CALCULUS  [§76 

Find  dy  in  Exercises  12-20. 

12.  y  =  tan  2xsin  2x.  15.  y  =  cos  (3  —  x)*. 

sin  2x 


14.  y  =  sin  (x^  +  3x  —  2).  17.  y  =  x  cos  2x  —  tan  2x. 

18.  y  =  tan-  (x  —  1). 

19.  y  =  cos*  (1  —  x»  —  2x). 

20.  2/  =  sin2  (2x  -  1)  cos^  (2x  -  1). 
Integrate : 

21.  dy  =  sin  2x<fx.  26.  dy  =  sin  x  cos  xdx. 

22.  dy  —  cos  2xdx.  26.  dy  =  tan  x  sec*  xdx. 

23.  dy  =  sec*  4x<ix.  27.  dy  =  Vsin  2x  cos  2xdx. 

24.  dy  =  sec  5x  tan  5xdx.  28.  dy  =  cos^x  sin  xdx. 

29.  dy  =  sec*  x  tan  xdx  =  sec'x  sec  x  tan  xdx. 

30.  dy  =  sec"  (x  —  1)  tan  (x  —  1)  dx. 

31.  Find  the  area  iinder  one  arch  of  the  sine  curve. 

32.  Find  the  area  under  one  arch  of  the  curve  y  =  2a*  sin*  x. 

1  —  cos  2x 
Hint,    sin*  x  = „ 

33.  The  equations  of  Exercise  8  are  the  parametric  equations  of  the 
cycloid.     Find  the  length  of  one  arch  of  the  cycloid. 

Hint,     ds  —  -\/{dxy  +  (dy)".     Express  ds  in  terms  of  e  and  dO. 

34.  Find  the  area  under  one  arch  of  the  cycloid. 

dv 
36.  X  =  a  cos  6,  y  =  a  sin  0.     Find  -j-.     Find   the  length   of    the 

curve.     Find  the  area  bounded  by  the  curve 

36.  x  = 

the  curve. 

dy 

37.  X  =  a  cos'  <j),  y  =  a  sin'  </>.  Find  -j-.     Find  the  length  of  the 

curve. 

38.  Find  the  volume  bounded  by  the  surface  obtained  by  revolv- 
ing y  =  sin  X  about  the  X-axis. 

39.  A  man  walks  at  the  constant  rate  of  4  feet  per  second  along  the 
diameter  of  a  semicircular  courtyard  whose  radius  is  50  feet.  The 
sun's  rays  are  perpendicular  to  the  diameter.  How  fast  is  the  man's 
shadow  moving  along  the  semicircular  wall  of  the  courtyard  when  he 
is  30  feet  from  the  end  of  the  diameter? 


dv 
36.  X  =  a  cos  e,  y  =  h  sin  6.     Find  t-.     Find  the  area  bounded  by 


§77]  CIRCULAR  FUNCTIONS  131 

40.  A  drawbridge  25  feet  long  is  raised  by  chains  attached  to  the  end 
of  the  bridge  and  passing  over  a  pulley  25  feet  above  the  hinge  of  the 
bridge.  The  chain  is  being  drawn  in  at  the  rate  of  6  feet  a  minute. 
Horizontal  rays  of  light  fall  on  the  bridge  and  it  casts  a  shadow  on  a 
vertical  wall.  How  fast  is  the  shadow  moving  up  the  wall  when  13 
feet  of  the  chain  have  been  drawn  in? 

41.  Find  ^  ii  x  =  y\/y  -  1. 

dy  .^ 


42.  Find  ^  ii  x  =  Vl  -  sin  y. 

43.  If  p'  =  a*  cos  2d  .  show  by  implicit  differentiation  that 
dp  _       a*  sin  29 

dd^  P 

44.  If  p  2  cos  0  =  a^  sin  36,  find  ^• 

46.    I  sin  6x  cos  2x  dx.  49.    I  sin  Ax  cos  7x  dx. 

46.  I   cos  4a;  cos  3x  dx.  60.    I  cos  5x  cos  9x  dx. 

47.  I   cos  5x  sin  2a;  dx.  61.    I  sin  wt  cos  at  dl. 

48.  I  sin  Sx  sin  3x  dx.  62.    1   cos  ut  cos  at  dt. 

63.  Find  the  mean  ordinate  of  the  curve  y  =  sin  x  between  the 
limits  X  =  0  and  x  =  tt. 

77.  Derivatives  of  the  Inverse  Circular  Functions. ^  The  for- 
mulas for  the  derivatives  of  the  inverse  circular  functions  are 
readily  obtained  from  those  of  §§75  and  76. 

.  1  The  student  will  recall  that  sin"'  u  is  defined  for  values  of  u  between  —1  and  +1 
only,  and  that  it  is  a  many  valued  function.  To  a  given  value  of  u  there  correspond 
infinitely  many  angles  whose  sines  are  equal  to  u.  This  will  be  seen  to  be  the  case  on 
sketching  the  curve  y  =  sin"'  u.  In  this  and  future  discussions  of  this  function  it 
will  be  made  single  valued  by  considering  only  those  values  of  y  =  sin"'  m  which  lie 

between  —  ^  and  +o'>  inclusive. 

The  positive  sign  of  the  radical  in  the  final  formula  (1)  is  chosen  because 
cos  V  ■»  \/l  —  uJ  is  positive  when  j/ lies  between  —^  and  +0"* 

Of  the  functions  occurring  in  (2),  (3),  (4),  (5),  and  (6),  y  =  cos"'  u,  and  y  =  sec"'  u 
are  made  single  valued  by  choosing  y  between  0  and  ir,  while  the  remaining  functions, 
y  —  tan"'  u,  y  =•  cot"'  u,  y  =  esc"'  u,  are  made  single  valued  by  choosing  y  between 

—  2"  and  +2"'     Show  that  the  proper  sign  has  been  chosen  for  the  radicals  in  the 

formulas  (2),  (5),  and  (6). 


132  CALCULUS  [§77 

Let  y  =  sin"^  u.     Then  sin  y  =  u.     Differentiation  gives 

dy       du 
cos  y  -r  =  T~' 
^  dx       dx 

dy  _      1     du 
dx  ~  cos  y  dx 

1 du 

Vl  —  sin"  y  dx 

du 
dy  dx 


Hence 


dx      y'l  _  u^ 


Therefore, 


du 


d(sin-iu)            dx  J/  .     ,    X            <*"              (1) 

— -— =      ,  »  or        d(sm-i  u)  =  —7- '^  ' 

The  student  will  show  that 

du 

d(cos-iu)               dx  J/        ,    X                du          (2) 

: =  —      ,  >      or       dtcos-^u)  = ,■  ^  ' 

dx  .              VI  -  u2  VI  -  u' 

du 

d(tan-»u)         dx  ^/*       ,    x          <^^                  (3) 
or      d(tan-'u)  =  f^p^z 


du 

or     d(cot-i  u)  =  -  YJ^i  (4) 


»  or     d(sec"^u)  =  7  -  (5) 

uVu*  -  1 


-  , ^     or     d(csc-i  u)  = ,  (6) 

dx  uVu"  -  1  uVu^  -  1 

Illustration  1.     K    y  =  sin"i(x^   —  2x   —   3),    find    dy.     By- 
formula  (1) 


dx 

"  1  +  u^' 

du 

d(cot-»  u) 
dx 

dx 

"     i  +  u*' 

du 

d(sec~^  u) 

dx 

dx 

u  Vu2  -  i 

du 

d(csc-'  u) 

dx 

§77]  CIRCULAR  FUNCTIONS  133 

d(x^  -2x-3) 


dy  = 


Vl  -  {x^  -2x-  3)« 
2(x  -  l)dx 


Vl  -  (x*  -2x-  sy 

Illustration  2.     U  y  =  tan~^  3x,  find  dy.     By  formula  (3) 

_     djSx) 
^y  ~  1  +  (3x)2 
_      3(Za; 
~  1  +  9x2" 

Illustration  3.     If  dt/  =  7-3^ — ^»  find  y. 


/r 


(ia; 


+  a;2 
or 

y  =  tan~^  x  +  C. 

Illustration  4.     li  dy  =  ^    ,  ^  ^y  find  r/. 

cix 


■'/r 


+  9x» 
3(ix 


+  (3x)2 

The  expression  under  the  integral  sign  is  now  of  the  form  -r-y — i 
whose  integral  is  tan~^  u.    Hence 

y  =  \  tan-i  (3x)  +  C. 

(XX 

Illustration  5.     If  dy  =  .    .   „  g>  find  ^. 


-/ 


dx 


4  +  9x2 
dx 


+  {ixy 


134  •        CALCULUS 

Hence 

y  =  i  tan-i  (ix)  +  C. 

Illustration  6.    If  dy  =  —    ^        ,  find  « 
V4  -  9x2  »• 


[§77 


9x5 


_  j^  r dx 

~  V  Vl-(|x)»' 

=  1.2  r ^dx 

'  VVi-(ix) 


The    expression    under   the   integral   sign  is  now  of  the  form 

du 
;^/p=|  whose  integral  is  sin-»  u.     Hence 

y  =  i  sin-i  (fx)  +  C. 


Exercises 

dy 
Find  ^  in  Exercises  1-10. 

1.  y  =  sin-i  (x^).  6.  2/  =  sin"'  (sin  x). 

2.  y  =  sin-i  (x  -  1).  f,  y  =  tan-i  -^— • 

X  —  1 

3.  2/  =  tan-Ux2).  8.  t/  =  sin-»  (1  -  x)K 

4.  y  =  tan-i  (x  -  1).  9.  y  =  gec-i  (a;2  _  3). 

5.  y  =  sm  (sin-»x).  10.  y  =  a;sin-ix. 
Integrate : 

11.  dy  =  t4%- 

^       4  +  x2 

12.  rfw  =  K-^^- 

^       9  +  x2 

13.  rfy  =         "^^ 


25  +  16x2 


-^     7     _        dx  It 

^  ~  a2  +  x^'  ^^'    y  =  a  *^"~'  ^  +  ^- 


CIRCULAR  FUNCTIONS  135 

dx 


Ans.    y  =  sin~*  — f-  C 


dx 


20.  dy  = .         —  = -•  Ans.    y  =  -  sec~i  -  +  C. 

Using  the  results  of  Exercises  14,  18,  and  20  as  formulas,  evaluate 
the  following  integrals: 


21. 
22. 
23. 
24. 
26. 
26. 
27. 


J  x/16^^^"  ^^'  J 
J  Vl6  -  9x»  J  ^ 

J  Vl  -  9x2  J 

J  25TT^  32.  J  - 

J  25  +  16x2*  33.  J 


dx 


x\/9x2  -  1 
dx 


X2  +  17 

dx 
Vl3  -  a;2 
dx 
\/x2  -  19' 
dx 


5x2  +  8 

dx 


J  1  +  16x2"  34-  J  ^2^43.4.5    J 

J  7v^2^^-  35.  J 


^   (x+2)2+l 
dx 


x\/3x2  -  14 


C  dx  «„      r      dx 

28.    I  -. 36.    I      , 

J    x\/9x2  -  25  J  \/9  -  x2 

37.  Find  the  area  between  the  ordinates  x  =  0,  x  =  §,  the  X-axis 

and  the  curve  y  =  — . 

Vl  -x2 

4a' 

38.  Find  the  area  under  the  curve  y  =    ,   ,    .   ,»  above  the  X-axis, 

"       x2  _j.  4o2  ' 


136 


CALCULUS 


[§78 


and  between  the  ordinates  x  =  0  and  x  =  b.     Find  the  limit  of  the 
area  as  b  increases  without  limit. 

39.  Find  the  mean  ordinate  of  the  curve  y  =  j--r — ^>  between  the 

■JT 

limits   X  =  0  and  x  =  -y 

78.  Velocity  and  Acceleration.     If  a  particle  is  moving  in  a 
curved  path,  its  velocity  at  any  point  is  represented  by  a  vector 

laid  off  along  the  tangent  with  its 
length  equal  to  the  magnitude  of 

ds 
the  velocity,  -r:'  Thus  the  ve- 
locity at  the  point  P,  Fig.  55,  is 
represented  by  the  vector  PT.  It 
can  be  resolved  into  the  com- 
ponents PK  and  PM,  parallel  to 
the  X-  and  F-axes,  respectively. 
~^  These  components  represent  the 
time  rates  of  change  of  the  coor- 
dinates of  the  moving  point  P,  i.e., 

PK^^"" 


Fig.  55. 


and 


Since 


PM  = 


dt 


dy 
dt' 


(1> 


PT  =  V{PK)^  +  {PM)\ 


ds 
dt 


(2) 


This  relation  can  be  obtained  directly  from  (2),  §63,  if  we  consider 
X  and  y  functions  of  t.  For,  we  can  divide  by  dt  and  obtain  the 
equation  (2). 

In  Fig.  55,  let  PT  be  the  velocity  at  P,  and  QT'  that  at  Q. 
Draw  from  a  common  origin,  o,  Fig.  56,  the  vectors  op  and  oq 
equal  to  the  vectors  PT  and  QT',  respectively.  Then  pq  equals 
the  vector  increment,   Av.     The   average   acceleration   for   the 

Av 
interval  At  is  equal  to  -rr  directed  along  pq.    Lay  off,  on  pq,  pm 


§78] 


CIRCULAR  FUNCTIONS 


137 


equal  to  ry    As  At  approaches  zero,  Q  approaches  P,  and  q  ap» 

proaches  p  as  indicated  by  the  dotted  line,  Fig.  56;  p7n  approaches  a 
vector  pt  directed  along  the  tangent  to  the  arc  pq  at  p.     This  vector, 

Av 
the  limit  of  — .•  represents  the  acceleration  of  the  particle  moving 

in  the  curved  path.    Let  us  calculate  its  x  and  y  components.     In 
Fig.  56,  denote: 

op  by  V  and  its  components  by  Vx  and  Vy, 

oq  by  v'  and  its  components  by  v'x  and  v'y, 

pq  by  Ay  and  its  components  by  Ay,  =  v'x  —  Vx  and  Ay„  =  v'„  —  Vy, 

pt  by  j  and  its  components  by  jx  and  j^. 


Fig.  56. 


Then 


Jx   = 


lim   Ay.  _  dv^ 

^'=^0    Af        dt 


lim  Ay^ 


dv„ 
dt 


dt 
dt 


d^ 
df 


d^ 
dt"^' 


The  magnitude  and  direction  of  the  vector  ^  are  given  by: 


(3) 


(4) 


(5) 


138 


CALCULUS 


[§79 


and 

where  0  is  the  angle  made  by  pt,  Fig,  56,  with  the  positive  direction 
of  the  X-axis. 

Again  we  can  resolve  the  acceleration  j  into  components  along 
the  tangent  and  normal.  In  Fig.  57,  PL  is  the  tangential  com- 
ponent and  PJ  is  the  normal  component.  The  tangential  com- 
ponent clearly  produces  the  change  in  the  magnitude  of  the 
velocity,  and  the  normal  component  the  change  in  its  direction. 


Fig.  57. 

79.  Angular  Velocity  and  Acceleration.  If  a  body  is  rotating 
about  an  axis,  the  amount  of  rotation  is  given  by  the  angle  d 
through  which  a  line  in  the  body  turns  which  intersects  the  axis 
and  is  perpendicular  to  it.  Thus  in  the  case  of  a  wheel  the  rotation 
is  measured  by  the  angle  6  through  which  a  spoke  turns.  ^  is  a 
function  of  the  time  t.  The  rotation  is  uniform  if  the  body  rotates 
through  equal  angles  in  equal  intervals  of  time.  If  the  uniform 
rate  of  rotation  is  co  radians  per  second,  the  body  rotates  through 
6  =  cat  radians  in  t  seconds.  If  the  rotation  is  not  uniform  the 
rate  at  which  the  body  is  rotating  at  any  instant,  the  angular 
velocity,  is 

<^^M^o^t      dt' 


§791  CIRCULAR  FUNCTIONS  139 

Similarly,  the  angular  acceleration  a  is  the  time  rate  of  change  of 
the  angular  velocity.     Then, 

do)      dW 
"~lt  ~  dt^' 

If  we  consider  a  particle  at  a  distance  r  from  the  axis  of  rotation, 
its  linear  velocity  v  is 

V  =  cor 

and  is  directed  along  the  tangent  to  the  circle  described  by  the 
particle.     The  tangential  acceleration  is 

jt  =  ar. 

Exercises 

,     1.  The  following  formulas  have  been  established  for  linear  motion, 
with  constant  acceleration : 

V    =   Vo  +  jt. 
S    =  Vot   +  ^jl\ 

I'  -  ^*  =  js.  (See  §38.) 

Show  that  the  corresponding  formulas  for  rotation  are: 

e  =  cjoi  +  W^' 

W^  COo^ 

T  -  "2-  =  «^- 

2.  A  flywheel  10  feet  in  diameter  makes  25  revolutions  a  minute. 
What  is  the  linear  velocity  of  a  point  on  the  rim? 

3.  Find  the  constant  acceleration,  such  as  the  retardation  caused 
by  a  brake,  which  would  bring  this  wheel  to  rest  in  30  seconds.  How 
many  revolutions  would  it  make  before  coming  to  rest? 

4.  A  resistance  retards  the  motion  of  a  wheel  at  the  rate  of  0.5 
radian  per  second  per  second.  If  the  wheel  is  running  at  the  rate  of  10 
revolutions  a  second  when  the  resistance  begins  to  act,  how  many 
revolutions  will  it  make  before  stopping? 

5.  A  wheel  of  radius  r  is  rotating  with  the  uniform  angular  velocity 
w.  Find  the  direction  and  magnitude  of  the  acceleration  of  a  point 
on  the  rim. 


140 


CALCULUS 


[§80 


Hint.  The  coordinates  of  the  point  can  be  written  x  =  r  cos  w<, 
y  =  rain.  uL     Find  -jtj  and  tt^* 

6.  A  wheel  of  radius  r  is  rolling  with  the  uniform  angular  velocity 
u  along  a  horizontal  surface  without  slipping.  How  fast  is  the  axle 
moving  forward?  The  parametric  equations  of  a  point  P  on  the  rim 
are: 

X  =  r{u}t  —  sin  cat) 
y  =  r(l  —  cos  (at). 

Find  the  magnitude  and  the  direction  of  the  velocity  of  P  at  any 
instant.  What  is  the  velocity  of  a  point  at  the  top  of  the  wheel? 
At  the  bottom? 

7.  If  a  particle  moves  in  such  a  way  that  its  coordinates  are 
re   =  a  cos  t   -\-  h,  y   =  a  sin  <  +  c,  where  t  denotes  time,   find  the 

equation  of  the  path  and  show  that  the  par- 
j>         tide  moves  with  constant  tangential  velocity. 

80.  Simple   Harmonic  Motion.    Let  the 

point  P,  Fig.  58,  move  upon  the  circumfer- 
ence of  a  circle  of  radius  a  feet  with  the 
uniform  velocity  of  v  feet  per  second,  so  that 

V 

the  radius  OP  rotates  at  the  rate  of  -  =  co 
Fig.  58.  o, 

radians  per  second.  The  projection,  B,  of  P 
on  the  vertical  diameter  moves  up  and  down.  If  the  point  P 
was  at  C  when  t  =  0,  the  displacement,  OB  =  y,  is  given  by 

y  =  asind  =  asinwf. 

If  the  point  P  was  at  D  when  t  =  0,  we  have 

y  =  asin(co^  —  a).  (1) 

Any  motion  such  that  the  displacement  at  time  t  is  given  by  (1) 
is  called  a  simple  harmonic  motion.  Thus  the  point  B,  Fig.  58, 
describes  simple  harmonic  motion.  The  abbreviation  "S.H.M." 
will  be  used  for  "simple  harmonic  motion." 

From  (1)  it  follows  that  the  velocity  of  a  point  describing 

S.H.M.  is 

dy 
dt 


(2) 


§81]  CIRCULAR  FUNCTIONS  141 

and  that  the  acceleration  is 

-77^  =  —a  cousin  (co^  —  a).  (3) 


The  second  member  is  —oihj,  by  equation  (1),     Hence 

-  a)2y,  (4) 


or 

df" 


+  co^y  =  0.  (5) 


Equation  (4)  shows  that  the  acceleration  of  a  particle  describing 
S.H.M.  is  proportional  to  the  displacement  and  oppositely  di- 
rected. That  the  acceleration  is  oppositely  directed  to  the 
displacement  is  to  be  expected  from  the  character  of  the  motion, 
which  is  an  oscillation  about  a  position  of  equilibrium.  Thus  if 
the  body  is  above  this  position  the  force  is  directed  downward, 
and  vice  versa.  In  Fig.  58,  the  point  B  has  a  positive  acceleration 
when  below  0  and  a  negative  acceleration  when  above  0.  The 
acceleration  is  zero  at  0,  a  maximum  at  the  lower  end  of  the  diame- 
ter, and  a  minimum  at  the  upper  end. 

In  accordance  with  (2)  the  velocity  is  zero  at  the  two  ends  of  the 
diameter.  The  velocity  has  its  greatest  numerical  value  when 
B  passes  through  0  in  either  direction. 

Equation  (4),  or  (5),  is  called  the  differential  equation  of 
S.H.M.     The   proportionality   factor  co^  is  connected  with  the 

period    T  by  the  relation   T  =  —      The    equation    (4)   was 

obtained  from  (1).  Frequently  it  is  desired  to  solve  the  converse 
problem,  viz.,  to  find  the  motion  of  a  particle  whose  acceleration  is 
proportional  to  the  displacement  and  oppositely  directed.  In 
other  words,  a  relation  between  y  and  t  is  sought  which  satisfies 
equation  (4).  Clearly  (1)  is  such  a  relation.  However,  it  will  be 
instructive  to  obtain  this  relation  directly  from  (4). 

First,  a  differential  equation  equivalent  to  (4)  will  be  obtained 
in  the  solution  of  the  problem  of  the  motion  of  the  simple  pendulum. 

81.  The  Simple  Pendulum.  Let  P,  Fig.  59,  be  a  position  of  the 
bob  of  a  simple  pendulum  at  a  given  instant  and  let  it  be  moving 
to  the  right.     If  s  denotes  the  displacement  considered  positive 


142 


CALCULUS 


[§81 


on  the  right  of  the  position  of  equiUbrium,  -tt^  is  the  acceleration 

ds 
in  the  direction  of  the  tangent  PT,  for  ^  is  the  velocity  along  the 

tangent.  This  acceleration  must  be  equal 
to  the  tangential  component  of  the  accelera- 
tion due  to  gravity,  if  the  resistance  of  the 
air  be  neglected.  This  component  is  equal  to 
—  g  sin  9.  Since  it  acts  in  a  direction  opposite 
to  that  in  which  s  is  increasing,  it  must  be 
taken  mth  the  negative  sign,  i.e.,  the  acceler- 
ation diminishes  the  velocity.     We  have  then 


Fia.  59. 


d's  .     . 


(1) 


If   the   angle  through    which   the    pendulum   swings    is    small, 
sin  6  can  be  replaced  by  6.     Then  (1)  becomes 

d^ 
dt^ 
Since  s  =  W, 


(2) 


d^ 
dV 


I 


(3) 


Putting  ,  =  w^  for  convenience  in  writing, 
d-'d 


dt^ 


=  -  co^d. 


dd 


Multiplying  by  2  -ir  and  integrating. 


\dt)     - 


oiW^  +  C». 


(4) 


The  arbitrary  constant  is  written  for  convenience  in  the  form  C. 
The  constant  must  be  positive.     Otherwise  the  velocity  would  be 
imaginary.    Extracting  the  square  root, 
dd 


dt 


=  VC*- w2  0»' 


or 


dd 


Vc^  -  w'  d^ 


=  dt. 


(5) 


§81]  CIRCULAR  FUNCTIONS  143 

Integration  gives 

—  sin~^  -p7   =  t  -\-  Ci. 

sin~^  -^  =  cof  +  coCi 

=  «<  +  C2, 
where  the  constant  coCi  is  replaced  by  the  constant  C2.     Then 
-T7  =  sin  (wf  +  C2), 

0  =  —  sin  (a>«  +  C2) 

CO 

=  C3  sin  {wt  +  C2), 
where  —  has  been  replaced  by  C3.     Therefore 

CO 

e  =  Ci  sin  (co<  +  C2)  (6) 

is  the  equation  of  the  angular  displacement  of  the  pendulum. 

2_.  /r 

The  form  of  (6)  shows  that  the  motion  is  of  period  —  =  2ir'\^-' 

It  is  a  S.H.M.  and  contains  two  arbitrary  constants.  They  can 
be  determined  by  two  conditions,  e.g.,  the  displacement  and 
velocity  at  a  given  instant.  Suppose  the  bob  drawn  aside  to  the 
right  so  that  the  string  makes  an  angle  9o  with  the  vertical.  The 
bob  is  then  released  without  being  given  an  impulse;  i.e.,  with  an 
initial  velocity  zero.  The  time  will  be  counted  from  the  instant 
of  release.    The  conditions  are  then 

d  =  do  (7) 

and 

f  =  »  («) 

when  t  =  6.     From  (6), 

do 

-jj  =  C0C3  cos  (cof  +  Ci). 

The  condition  (8)  gives 

0  =  C0C3  cos  C2, 

or  cos  C2  =  0.     Whence  C2  =  ^-     Then  (6)  becomes 
6  =  Cz  cos  cot. 


144  CALCULUS  [§81 

The  condition  (7)  gives 

do  =  Cz. 
Hence 

e  =  do  cos  oit.  (9) 

Multiplying  by  I  and  recalling  that  W  =  s,  and  denoting  Ido  by 
So,  we  have  as  the  equation  for  the  displacement  s, 

S  =  So  cos  bit.  (10) 

The  period  is  T  =  —  =  27r  'y-.    When  is  the  velocity  of  the 

bob  greatest?    When  least,  numerically? 

Equation  (6),  the  solution  of  (4),  shows  that,  if  the  acceleration 
of  a  particle  is  proportional  to  its  displacement  and  oppositely 
directed,  the  particle  describes  S.H.M. 

Exercises 

1.  Write  the  differential  equations  of  the  following  simple  harmonio 
motions.     Find  the  period  in  each  case. 

y  =  5  sin  3^ 

2/  =  6  sin  Izt  +  Ij  • 

y  =  5  cos  3/. 

2/  =  4  sin  2t  +  S  cos  2t. 
J/  =  7  sin  {8t  +  a). 

2.  Write  the  equation  of  a  S.H.M.  which  satisfies  the  equations: 

J  +  3!,-0. 


CHAPTER  IX 

EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 

82.  Derivative  of  the  Exponential  and  Logarithmic  Functions. 
Let 

T/  =  a*.  (1) 

Then 

Ay  =  a'(a^  -  1) 
Ay  _  ^/(i^  —  1\ 
Ax~       \     Ax     ) 

dx~  ^   Ax^o      Aa;  ^'^> 

„,        a^  —  I  .    .    .  .  ,        lim  fl"^  —  1  . 

bince  — -r IS  independent  of  x,   ^^q  — r is  a  constant 

for  a  given  value  of  a.     Call  this  constant  K,  so  that 
Then  from  (2), 


_    lim  «^-  1  ,^. 


%  =  Ka:  (4, 

Equation  (4)  shows  that  the  slope  of  the  curve  y  =  a''  is  propor- 
tional to  the  ordinate  of  the  curve.  In  other  words,  the  rate  of 
increase  of  the  exponential  function  is  proportional  to  the  function 
itself. 

When  a;  =  0,  it  follows  from  (2)  and  (3)  that 

dy         Hm   a^  -  1 


dx       ^-0       Ax 


=  K. 


Consequently  the  constant  K  introduced  above  is  the  slope  of  the 
curve  t/  =  a^  at  the  point  (0,  1).  This  slope  depends  upon  the 
value  of  a.  Let  e  be  that  particular  value  of  a  for  which  the 
corresponding  curve,  y  =  e^,  has  a  slope  equal  to  1  at  the  point 
where  it  crosses  the  F-axis. 

10  145 


146  CALCULUS  [§82 

If,  then, 


equation  (4)  becomes 


y  =  6"=,  (6) 

dy 


dx 


since  K  =  \'\n  this  case.     Or 

de 


Then 
and 


dx  =''•  (6) 

de»  du  _. 


de»  =  cdu.  (8) 


Equation  (6)  shows  that  the  slope  of  the  curve  y  =  e'  is  equal 
to  the  ordinate  of  the  curve.  The  number  e  is  the  base  of  the 
natural  system  of  logarithms.  It  is  sometimes  called  the  Naperian 
base.  Its  value,  2.71828  •  •  •  •  ,  will  be  calculated  later  in 
the  course. 

The  formula  for  the  derivative  of  the  natural  logarithm  of  a 
function  is  now  easily  obtained.  In  calculus  if  no  base  is  indi- 
cated, the  natural  base  is  understood.    Thus  log  u  means  log,  u. 

If 

y  =-  log  u, 
u  =  e« 


du  dy 


and  by  (7) 
Whence 


That  is 

or  (9) 


dx  '' 

=  ^"d|- 

dy 
dx  ' 

_  1  du 
~  e"  dx 

= 

_  1  du 
~  u  dx 

d(log 

u)  _  1  du 

dx 

~  u  dx' 

d(Iog  1 

du 

u)  =  — • 
'       u 

§82]   EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    147 

Since  ^ 

logoW  =  logoclogw,  (10) 

d  (logau)       ,  „  ^  1  du 

or  (11) 

du 

d(logaU)   =   logaC  — ' 

11  y  =  a",  log  y  =  uloga 

l  dy  _  ,         du 
y  dx  ~  dx 

dy         .         du 

Tx^^y^^'^^'Tx 

.         du 
=  aMoga^. 


That  is 


da"  ,        du 

^  =  auloga^ 

da«  =  a«»  log  a  du 


(12) 


Illustrations. 

1.  If  2/  =  e*',  dy  =  e^'d(x2)  =  2xe''"dx. 

2.  If  2/  =  e''"" ",  dy  =  e  "''^  ''d{smx)  =  cos  x  e"°  *  dx. 

r.       Tr  1  /  .      HX  1  1  ^(^  +   1)  1  dx 

3.Uy  =  logio  (x  +  1),  dy  =  logio  e    ^  ,   ^     =  logio  e^-^y 


•l.    J.1   J/    =    lUJ 

U^  t 

-  ^), 

t 

''V 

x  +  V 

»Let 

z 

= 

log  u 

Then 

& 

= 

u 

Taking  logarithms 

to  the 

base 

1, 

z 

log 

a  e 

= 

logo  U 

That  is 

log 

u  1 

og» 

1  e 

=    logo  M 

148  CALCULUS  [§82 

5.  Uy  =  e*^'^"'^  dy  =  e*'*'^'"^ d(tan-i x)  =  e*^"" '^ rr^ • 

1  -\-x 

6.1iy  =  log  1^^,  y  =  2  log  (1  +  x)  -  3  log  (1  -  X), 

and 

,  2dT;     ,     Sdx         5  +  X    , 

7.  If  2/  =  e*  sin  x,  -r-  =  e*(cos  x  +  sin  x) 


and 


-j-^  =  2e*  cos  X. 


Exercises 


Find  the  first  derivative  of  each  of  the  following : 

1.  y  =  e'*.  6.  2/  =  log  (1  -  x^).       11.  y  =  e^''^*. 

2.  y  =  e'\  7.  y  =  e'  cos  x.  12.  y  =  e***^  ^'. 

Z.  y  =  log  (x").  8.  2/  =  e'».  13.  y  =  log  Vx''  -  1. 

4.  y  =  log  (x').  9.  y  =  e'=^',  14.  y  =  e~*  sin  x. 

6.  y  =  log  (x*  -  1).     10.  y  =  log  [^^^I'      16-  2/  =  10'- 

16.  y  =  logio  X. 

17.  y  =  5'. 

18.  y  =  x^  5*. 

19.  Show  that  the  aubtangent  for  the  curve  y  =  a''  is  a  constant. 
What  is  this  length  when  a  =  e? 

Illustrations. 

8.  If  dy  =  e'rfx,       y  =    \  e'dx  =  e'  -\-  C. 

9.  If  dy  =  xe'^dx. 

y  =   \  xe'^dx 

=  |e''  +  C. 
10.  If  dy  =  ^> 

X 

J/  =  log  X  +  C 
=  log  X  +  log  i2! 
=  log  Kx. 


§82]   EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    149 

dx 

11.  If  dy  =  ^:^, 

y  =  \og{x  +  l)  +  C 
=  log  K(x  +  1). 

xdx 

12.  If  dy  =  ^^^p3, 


a;2  +  l 
2xdx 


x'  +  l 
=  i  log  (x^  +  1)  +  (7 
=  log  Vx2  +  1  +  C 
=  log  KVx^  +  1. 

13.  If  dy  =  e™*  cos  xdx, 

y  =  j  e  "°  *  COS  X  dx 

1.1    Tf  J  (x  +  l)dx 

14.  If  dy  =  ^2  +  2x  +  3' 


-/ 


(x  +  l)dx 


x2  +  2x  +  3 

r  2(x  +  l)dx 
■^  ^  J  x2  +  2x  +  3 

=  h  log  (x2  +  2x  +  3)  +  (7 

=  log  Vx^  +  2x  +  3  +  C 

=  log  KVx^  +  2x  +  3. 

15.  If  dy  =  tan  x  dx, 

y  =   I  tan  x  dx 


-/- 


sin  X  , 

ax 

cosx 


=  —  log  cos  X  +  (7 
=  log  sec  X  +  C. 


150 


CALCULUS 


l§82 


16.  If 


17.  If 


18.  Find 
Let 

whence 
and 


Then,  since 


dy  =  cot  xdx, 
y  =    \  cot  X  dx 


Si 


dx 


sin  X 
=  log  sin  X  -^  C. 
dy  =  sec  x  dx, 
y  =  J  secx  dx  -\-  C 

(sec  x  +  tan  x)  sec  xdx 
sec  X  +  tan  x 

/-■'■■""•" 


f- 


dx 


sec  X  -J-  tan  x 
=  log  (sec  X  +  tan  x)  +  C 
dx 


S 


Vx^  ±  a2 


v^  =  x^  ±  a" 
2v  dv  =  2x  dx 


dv 

X 

dv 

X 


dx 

dx 

V 


dx         _    Ci 


dx  +  dv 
X  +  V  ' 

dx  +  c^p 


19.  If 


Vx^  ±  a2       J    X  +  v 

=  log  (x  +  v)  -\-C 

=  log  (a;  +  Vx*  ±  a2)  +  C. 


y 


■/. 


e'  +  1 
=  log  (e*  +  1)  +  C. 


EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    151 

20.  If  ^^  =  ^, 

y      X 

log  2/  =  log  X  +  log  C 
log  y  =  log  Cx 
y  =  Cx. 

21.  If  —  =  n  —  J 

y  X 

log  y  =  n  log  a;  +  log  C 
=  log  a;»  +  log  C 
=  log  Cx", 
1/  =  Cx". 

Exercises 

The  results  of  Illustrations  15,  16,  17,  and  18  are  to  be  used  as 
formulas  of  integration. 

In  the  following  exercises  and  y : 

20.  dy  =  x2  e*'  dx.  27.  dy  =  tan  2x  dx. 

21.  dy  =  e*-""  sec"^  X  dx.  28.  dy  =  cot2xrfx. 

22.  dy  =  — r-r.  29.  dy  =  sec  2x  dx. 

"       X  +  1  " 

23.  dy  =  Y^x  ^°-  "^^  =  (logx)*^. 

_,     -  xdx  «^     ,  cos  X  dx 

2*-  ^^  =  IT^^'  31.  dy  =  ^-j^^^^^- 

26.  d2/  =  ^,-  32.  d2/  =  (!l-ZiZM£. 

'^       1  —  x^  -^  e*  +  e  * 

«/.     ,  (e"^  +  l)dx  „„     ,         sec^xdx 

(e2*  -  l)dx 


34.  dy  = 


+  1 


35.  Find  the  area  between  the  equilateral  hyperbola  xy  =  10,  the 
X-axis,  and  the  lines  x  =  1  and  x  =  2. 

36.  The  volume  of  a  gas  in  a  cylinder  of  cross  section  A  expands 
from  volume  vi  to  volume  Vi.  If  it  expands  without  change  in  tem- 
perature the  pressure,  p,  on  the  piston  varies  inversely  as  the  volume 
(Boyle's  Law,  pv  =  K).    Show  that  the  work  done  by  the  expansion  is 


pdv 


K\  ?  =  xiog^;. 


152  CALCULUS  [§82 

37.  The  subtangent  of  a  curve  is  of   constant  length,  k.     Find  the 
equation  of  the  curve. 

38.  For  what  value  of  x  is  the  rate  of  change  of  logio  x  the  same 
as  the  rate  of  change  of  z? 


-I 

I  tan  36  dd.  48.     | 


39.    I  CSC  X  dx.     (See  Illustration  17.) 

xdx 


41.  j  e*"'^''  sin  e  dd.  49.    j  lO'dx. 

42.  I  cot  I  dd.  60.    I 


cos^  (5x  —  4) 
2    (1  —  Bmx)dx 


43.  j  X  sec2(z2  +  l)dx.  51.    | 
I                         '  I  a;  +  cos  X 

44.  I  sec  I  dx.  52.    j  e^^  +  """^  ">  sin  0  dd. 

Cdx  C  xHx 

46.    I    —  63.    I    ;  ,     3- 

46.    f'-^^.  64.    f-^:^. 
J„  Vl+x^  J  Va^  +  X* 

66.    I  (tan  20  —  \)^de  =  \  tan  20  +  log  cos  20  +  C. 


/ 

/Idx 


57.    I  C8c(7x  +  5)  dx. 

68.    1-^+^)^" 

Vx^  +  4x  +  7 

69. 

~  (3x  +  2)dx 
"•    '  3x*  +  4x  +  9 


§83]   EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    153 

(x  +  2)dx 


61 


/. 


(x"  +  4x  +  7)2 

62.  Show  that  y  =  ae~**  cos  o)t  satisfies  the  differential  equation 

g+2fc*  +  („.+*=)„-0. 

63.  Find  the  mean  ordinate  of  the  curve  w  =  -  between  the  limits 

"         X 

X  =  1  and  X  =  2. 

83.  Logarithinic  Differentiation.  It  is  often  advantageous  in 
finding  the  derivative  oi  y  =  f{x)  to  take  the  logarithm  of  each 
member  before  differentiating.  A  number  of  examples  will  be 
solved  to  illustrate  the  process. 


(x-  1)3 


lUllSLIUUUII,    1. 

X'  lllU 

Liie  ut 

{x  +  1)^ 

y  = 

{x  -  1)^ 
(x+  1)' 

and  take  the  logarithm 

of  each  member. 

log  3 

!/  =  1  log  (X 

-  1)  -  1  log  (x  +  1). 

Differentiating, 

Idy 
y  dx 

2                    3 

3(x 

-  1)       5(x  +  l) 

Let 


x  +  19 


15  (a;2  -  1) 
dy  _     g  +  19 
dx  ~  15  (a;2  -  1) 


2/ 


X  +  19      (x  -  1)^ 


15  (x2-  1)  (x  +  1)' 
x  +  19 

15  (x  -  l)^  (x  +  1)^ 

1 11.11  RtTniinti.  2 

•\/l  —  x^ 
Find  the  derivative  of ■" 

i/x^  +  i 

Vl-a;^ 

2/  -     3/-^^^^ 

154  CALCULUS  l§83 

log  2/  =  §  log  (1  —  x2)  —  A  log  (x2  4-  1) 
1  dy  X  2x 

ydx^~  1  -x2  ~  3  (a;2  +  1) 

_  _  3^  (5  +  a:") 
~       3  (1  -  x^) 
dy a;  (5  +  a:')  Vl  -  x^ 

dx  ~    3(1 -x^)  v^mh: 

a:  (5  +  a;2) 


3Vl  -a;2(x2+  1)^ 

This  method  is  manifestly  shorter  and  simpler  than  that  of  differ- 
entiating by  the  rule  for  the  derivative  of  a  quotient. 
Illustration  3.     Find  the  derivative  of  {x^  +  l)''''^^. 
y  =  {x"^  -\-  1)3^^2 
log  y  =  (3a;  +  2)  log  {x^  +  1) 


dy 


J^=(^^  +  2)^-^  +  3log(x^  +  l) 


^^  =  [(3a:  +  2)  ^^  +  3  log  (x2  +  1)]  {x^  +  1)3-+^ 

-  -p  is  called  the  logarithmic  derivative  of  y  with  respect  to  x. 
It  will  be  considered  further  in  a  later  article. 

Exercises 
Find  the  derivative  in  Exercises  1-8. 

3 

1.  2/  =  ^^^tilL-  3.  2/  =  (X  +  1)3  (2x  +  5)1 

(x-7)5 

2-  ^  =  (x-4)t(r5)^-  *•  ^  =  ^(1  +  ^)^^^- 

5.  2/  =  x"n^.     (Solve  by  two  methods.) 

6.  7/  =  x^^°^. 

7.  s  =  (7<  +  3)10^-2. 

8.  2/  =  xV'^. 

In  Exercises  9-16  find  the  logarithmic  derivative. 

9.  2/  =  e^'.  12.  y  =  x". 
10.  y  =  x\                                        13.  y  =  ex". 


§84]   EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    155 

11.  y  =  Zi'  14.  2/  =  e*'+^  =  ce**. 

15.  y  =  10*'+^. 

16.  y  =  uv,  where  u  and  v  are  functions  of  x. 

17.  y  =  uvw,  where  u,  v,  and  w  are  functions  of  x.     Find  t— 
Find  y  if  its  logarithmic  derivative  is: 


22.  A;. 

(1) 

(2) 

Equation  (2)  expresses  the  fact  already  noted  in  §82,  as  a 
characteristic  property  of  the  exponential  function,  viz.,  that  the 
function  increases  at  a  rate  proportional  to  itself.  We  can  show, 
conversely,  that  if  a  function  increases  at  a  rate  proportional  to 
itself,  it  is  an  exponential  function. 

Thus,  let  it  be  given  that 
dy 


18.  61  +  7. 

Ans.  y  =  Ce  ' 

X 

19  J. 

X 

20.  - 

X 

21. 

FkxY 

84. 

Compound  Interest  Law.     If 

y  = 

Ce*S 

dy 
dt 

CA;e*' 

=  ky. 

dt  =  ^y-  (3> 


Then 

dy 


kdt 


logy  =  kt  +  C 

2/  =  e*'+c  =  e^e*'. 
Hence 

y  =  Cie*'.  (4) 

When  a  function  varies  according  to  this  law  it  is  said  to  follow 
the  "compound  interest  law."  For,  if  a  sum  of  money  be  placed 
at  compound  interest,  its  rate  of  increase,  for  any  interest  term,  is 
proportional  to  the  amount  accumulated  at  the  beginning  of  that 
term.  The  more  frequently  the  interest  is  compounded  the 
more  nearly  does  the  amount  accumulated  increase  according 
to  the  exponential  law. 

In  many  cases  in  nature  the  function  decreases  at  a  rate  pro- 


156  CALCULUS 

portional  to  itself.     The  compound  interest  law  appears  in  this 
case  in  the  form  Ce~*',  where  A  is  a  positive  constant.     For,  if 

^=  -kv 

it  follows  that 

y  =  Ce-*«. 

Illustration  1.  Newton's  law  of  cooling  states  that  the  tem- 
perature of  a  heated  body  surrounded  by  a  medium  of  constant 
temperature  decreases  at  a  rate  proportional  to  the  difference  in 
temperature  between  the  body  and  the  medium.  Let  d  denote  the 
difference  in  temperature.     Then 

ft  =  -  ^^-  (S) 

and 

e  =  Ce-*«.  (6) 

The  meaning  of  the  constant  C  is  seen  at  once  on  setting  t  =  0. 
It  is  the  difference  in  temperature  between  the  body  and  the 
medium  at  the  time  t  =  0.  If  this  initial  difference  in  temperature 
is  known,  (6)  gives  the  temperature  of  the  body  at  any  later 
instant.  Call  the  initial  difference  in  temperature  ^o-  (6) 
becomes 

e  =  doe-"'.  (7) 

The  time  which  is  required  for  the  difference  in  temperature  to 
fall  from  di  to  62  can  be  found  from  (7).     Thus 
01  =  0oe-*'i 
62  =  ^oe-*'» 


whence 


/.-<.  =  ^  log  |.  (8) 

This  result  could  have  been  obtained  directly  from  the  differ- 
ential equation  (5).     Thus 

..=  -if.  (9) 


§84]   EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    157 


Integrating  the  left-hand  member  between  the  limits  ti  and  <2  and 
the  right-hand  member  between  the  limits  di  and  62, 


f— iX 


e 


Illustration  2.     Find  the  law  of  variation  of  the  atmospheric 
pressure  with  height. 

Consider  a  column  of  air  of  unit 
cross  section  (Fig.  60).  Denote 
height  above  sea  level  by  h  and  the 
pressure  on  unit  cross  section  at  this 
height  by  p.  The  difference  in  pres- 
sure at  C  and  D  is  the  weight  of  the 
gas  within  the  element  of  volume 
of  height  Ah. 
Thus 

Ap  =  —  gp  Ah, 

where  p  is  the  average  density  of  the  air  in  the  volume  CDEF. 
Then 


Fig.  60. 


and 


Ap 

Ah 

dp 
dh 


=  -  gp 


=  -  gp, 


(1) 


where  p  is  the  density  at  C.  If  the  temperature  is  assumed  con- 
stant, the  air  obeys  Boyle's  Law,  pv  =  c,  where  v  denotes  the 
volume  occupied  bj'^  unit  mass  of  air.     Since 

mass     _  1^  _  P 
volume       V        c 


P  = 


dh 


=  -  kp, 


158 

CALCULUS 

where 

*  =  ^ 

Integration  gives 

[§84 


logp  =  —  kh  +  logCi, 
or 

When  h  =  0,  p  =  Po,  the  pressure  at  sea  level,  and  Ci  =  po- 

Hence 

p  =  poe~**.  (2) 

If  h  is  measured  in  meters  and  p  in   millimeters  of  mercury, 
k  =  risVir,     (2)  becomes 

p  =  760e-iToT-  (3) 

Exercises 

1.  A  law  for  the  velocity  of  chemical  reactions  states  that  the 
amount  of  chemical  change  per  unit  of  time  is  proportional  to  the 
mass  of  changing  substance  present  in  the  system.  The  rate  at  which 
the  change  takes  place  is  proportional  to  the  mass  of  the  substance 
still  unchanged.  If  q  denotes  the  original  mass,  find  an  expression 
for  the  mass  remaining  unchanged  after  a  time  t  has  elapsed. 

2.  Assuming  that  the  retardation  of  a  boat  moving  in  still  water  is 
proportional  to  the  velocity,  find  the  distance  passed  over  in  time  t 
after  the  engine  was  shut  off,  if  the  boat  was  moving  at  the  rate  of  7 

7 
miles  per  hour  at  that  time.     Ans.  s  =  t(1  —  e  *'). 

3.  The  number  of  bacteria  per  cubic  centimeter  of  culture  increases 
under  proper  conditions  at  a  rate  proportional  to  the  number  present. 
Find  an  expression  for  the  number  present  at  the  end  of  time  t.  Find 
the  time  required  for  the  number  per  cubic  centimeter  to  increase 
from  6i  to  62-  Does  this  time  depend  on  the  number  present  at  the 
time  t  =  0? 

4.  A  disk  is  rotating  about  a  vertical  axis  in  a  liquid.  If  the  retar- 
dation due  to  friction  of  the  liquid  is  proportional  to  the  angular 
velocity  «,  find  w  after  t  seconds  if  the  initial  angular  velocity  was  wo. 

6.  If  the  disk  of  Exercise  4  is  rotating  very  rapidly,  the  retardation 
is  proportional  to  w^  Find  w  after  t  seconds  if  the  initial  angular 
velocity  was  wq. 


§85]   EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    159 

85.  Relative  Rate  of  Increase.  If  the  rate  of  change  of  a  func- 
tion is  divided  by  the  function  itself,  the  quotient  is  the  rate  of 
change  of  the  function  per  unit  value  of  the  function.  This 
quotient  has  been  called  the  relative  rate  of  increase  of  the  function. 
If  a  function  varies  according  to  the  compound  interest  law,  its 
relative  rate  of  increase  is  constant,  i.e., 

y  dt 

One  hundred  times  the  relative  rate  of  increase  is  the  percent  rate 
of  increase.     Thus  if 

-  t  =  0.02, 

V  dt  ' 

the  percent  rate  of  increase  is  2.  This  means  that  y  increases  2 
percent  per  unit  time.  Any  of  the  Exercises  1-5  might  have  been 
stated  in  terms  of  the  relative  rate  of  increase  of  the  function 
concerned. 

Exercises 

1.  Given  that  the  intensity  of  light  is  diminished  2  percent  by 
passing  through  one  millimeter  of  glass,  find  the  intensity  /  as  a  func- 
tion of  t,  the  thickness  of  the  glass  through  which  the  light  passes. 

2.  The  temperature  of  a  body  cooling  according  to  Newton's  Law 
fell  from  30°  to  18°  in  6  minutes.  Find  the  percent  rate  of  decrease  of 
temperature  per  minute. 

86.  Hyperbolic  Functions.  The  engineering  student  is  likely 
to  meet  in  his  reading  certain  functions  called  the  hyperbolic 
functions.  These  functions  present  analogies  with  the  circular 
functions  and  they  are  called  hyperbolic  sine,  written  sinh,  hyper- 
bolic cosine,  written  cosh,  and  so  on. 

These  functions  are  defined  by  the  equations: 

cosh  X  =  — n — '       sech  x  = 

.  ,           e'  —  e~'  , 

smh  X  =  — -X »        csch  x  = 


2 

gx  —  g-x 

6'  —  e~* 


tanh  X  =  ..  ,   ,_.'       coth  x    = 
odd  functions 


e*  4-  e~*  tanh  x 

cosh  X  and  sech  x  are  even  functions,  while  the  remaining  four  are 


160  CALCULUS  [§87 

Exercises 

1.  By  making  use  of  the  definitions  the  student  will  show  that  the 
following  identities  hold.  They  are  analogous  to  those  satisfied  by  the 
circular  functions. 

cosh^x  —  sinh^x  =  1. 
1  —  tanh*x  =  sech^x. 

2.  Show  by  the  use  of  the  defining  equations  that : 

d  cosh  X 


dx 
d  sinh  X 

dx 
d tanh  x 

dx 
d  coth  X 

dx 
d  sech  X 

dx 
d  csch  X 


sinh  X, 
=  cosh  X. 
=  sech^  X. 
=  —  csch^  X. 
=  —  sech  X  tanh  x. 
=  —  csch  X  coth  X. 


dx 

3.  Sketch  the  curves  y  =  cosh  x,  y  =  sinh  x,  and  y  =  tanh  x. 

87.  Inverse  Hyperbolic  Functions.     The  logarithms  of  certain 
functions  can  be  expressed  in  terms  of  inverse  hyperbolic  functions 

Let 

y  =  sinh~*  x. 

e«  —  e~y 


J/  —  Buiii  (/  —        2 

or 

e'^y  —  2xey  —  1=0, 

whence 

c  =  X  ±  Va;^  +  1. 

The  minus  sign  cannot  be  taken  since  e"  is  always  positive. 
Hence 

e"  =  a;  +  Vx^  +  1  , 
and 

y  =  sinh-'  x  =  log  (x  +  Va;^  +  !)• 


1.  Show  that 
Since 


§88]   EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    161 

Exercises 

cosh~i  a;  =  log  (x  +  va;"  —  1). 

X  -  -y/x^  -  1  =  7 > 

X  +  Vx'^  -  1 

log  (x  -  a/x"  -  1)  =  -  log  {x  +  \/x2  -  1). 
Therefore 

cosh~ia;  =  +  log  (x  +  's/x^  —  1). 

The  inverse  hyperbolic  cosine  is  then  not  single  valued.     Two  values 

of  cosh~i  x,  equal  numerically  but  of  opposite  sign,  correspond  to  each 

value  of  x  greater  than  1. 

2.  Show  that: 

1  +  x 
tanh'^x  =  §  log  ,  _    >  if    x'^  <  1; 

X  +  1 
coth~i  X  =  2  log      _  ^»  if     x^  >  1 ; 

1    -f.    a/1    _  a;2 

sech~*  X  =  ±  log >     if     0  <  x  ^  1 ; 


and 


,  _.  ,       1  +  Vx^  +  1  .. 

csch  ^x  =  log >  if     X  >  0: 

^  X  ' 


csch  ^  X  =  log >  if     X  <  0. 


The  student  is  not  advised  to  memorize  the  formulas  of  this  and  the 
preceding  sections  at  this  stage  in  his  course,  but  to  acquire  sufficient 
familiarity  with  the  hyperbolic  functions  to  enable  him  to  operate  with 
these  functions  by  referring  to  the  definitions  and  formulas  given  here 
and  to  others  that  he  will  find  in  mathematical  tables. 

88.  The  Catenary.  Let  AOB,  Fig.  61  a,  be  a  cable  suspended 
from  the  points  A  and  B  and  carrying  only  its  own  weight.  Let 
us  find  the  equation  of  the  curve  assumed  by  the  cable,  consider- 
ing it  homogeneous.  We  shall  assume  that  the  curve  has  a 
vertical  line  of  symmetry,  OY,  and  that  the  tangent  line  drawn 
to  the  curve  at  0  is  horizontal. 

Take  07  as  the  F-axis.  Imagine  a  portion  of  the  curve,  OP, 
of  length  s,  cut  free.  To  hold  this  portion  in  equilibrium  the 
forces  //  and  T,  Fig.  61  h,  must  be  introduced  at  the  cut  ends. 
n 


162 


CALCULUS 


H  and  T  are,  respectively,  equal  to  the  tension  in  the  cable  at 
the  points  0  and  P  and  they  act  in  the  direction  of  the  tangent 
lines  drawn  to  the  curve  at  these  points.  The  portion  of  the  cable 
OP,  Fig.  61  b,  is  in  equilibrium.  Hence  H',  the  horizontal  com- 
ponent of  T,  is  equal  to  H. 


V,  the  vertical  component  of  T,  must  balance  the  weight  of  the 
portion  OP  of  the  cable.     Hence 

V  =  sw, 
where  w  is  the  weight  of  a  unit  length  of  the  cable. 
From  Fig.  61  6,  it  is  seen  that 

dy  _   V^  _  V^  _  ws 
dx~  H'  ~  H  ~  H' 
Let 

w  _  1 
H  ~  a 
Then 

dy       s 

di  "  a'  (^) 

This  differential  equation  involves  three  variables,  viz.,  x,  y,  and 
8.  s  may  be  eliminated  by  differentiating  and  substituting  for  ^ 
its  value, 


Thus 


dx 


a  dx       a  \^  +  W 


EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS    163 
The  equation  now  involves  only  two  variables  and  may  be  written 


,  dy 

ok  upon  -j- 

tion  (2)  is 


=  -dx.  (2) 


dv 
If  we  look  upon  -r-  as  the  variable  u,  the  left-hand  side  of  equa- 


du 


vTTw2 

whose  integral  is  log  (u  +  \/l  -|-  u^).     (See  Illustration  18,  §82.) 
Integrating  (2), 

When  a;  =  0,  ^  =  0. 

Hence  C  =  0  and  (3)  becomes 


dv 
From  the  symmetry  of  the  curve  -r-  changes  sign  when  x  is  re- 
placed by  —X.     Then  from  (4), 

Subtracting  (5)  from  (4), 

2g=e«-e"a,  (6) 

or 

^  =  sinh?  (7) 

dx  a 

Integrating  (7), 

y  =  a  cosh  -  -f-  d-  (8) 

If  the  origin  is  taken  a  units  below  the  point  0,  Fig.  61  a, 
y  =  a  when  x  =  0,  and  C2  =  0.     Hence 

y  =  a  cosh  —  (9) 


164  CALCULUS 

This  is  the  equation  of  the  curve  assumed  by  the  cable.  It 
is  called  the  catenary. 

Equation  (9)  can  be  written 

Y  =  cosh  X,  (10) 

where 

r  =  ^      and      X  =  ?• 
a  a 

The  constant  a  depends  upon  the  tautness  of  the  cable.  Equa- 
tion (10)  shows  that  the  curve  y  =  cosh  z  if  magnified  the  proper 
number  of  diameters  will  fit  any  cable  hanging  under  its  own 
weight. 

The  length  of  OP  can  be  found  by  substituting  in  formula  2,  §63, 

du 
the  value  of  j-  given  by  (7),  and  integrating. 


-^' 


ds   =  «  / 1  +  sinh'  -  dx 
a 


1C 

=  cosh-  dx. 
a 

X 

s  =  a  sinh     +  C3. 
a 

Since  s  is  measured  from  the  point  where  the  curve  crosses  the 
y-axis,  s  =  0  when  x  =  0.     Hence  Cj  =  0  and 

s=  a  sinh  -•  (11) 

a 

Exercises 

1.  If  the  two  supports  A  and  B,  Fig.  61,  are  on  a  level,  L  feet  apart, 
and  if  the  sag  is  d  feet,  show  that  the  tension,  T,  in  the  cable  at  the 
points  A  and  B  is 

T  =  w(a  +  d). 

2.  Beginning  with  equation  (6)  find  expressions  for  y  and  s  without 
making  use  of  hyperbolic  functions. 

3.  If  the  cable.  Fig.  61,  is  drawn  very  taut,  show  that  the  equation 
of  its  curve  is  approximately 

X2 


y  = 


2a 


if  the  origin  of  coordinates  is  taken  at  the  lowest  point  of  the  cable. 

Hint.     Begin  with  equation  (2)  and  note  that  i-r-j  is  small  com- 
pared with  1. 


CHAPTER  X 
MAXIMA  AND  MINIMA 

In  previous  chapters  maximum  and  minimum  values  of  func- 
tions have  been  found  by  making  use  of  the  derivative.  Besides 
this  method  several  others  which  do  not  involve  the  use  of  the 
derivative  may,  at  times,  be  used  to  advantage. 

89.  The  Maximum  or  Minimum  of  y  =  ax^  +  j8x  +  7.  In 
elementary  analysis  the  student  learned  that  y  =  az^  +  j8a;  +  7 
represents  a  parabola  with  its  axis  parallel  to  the  F-axis,  and 
that  the  equation  can  be  put  in  the  form  y  =  a(x  —  p)^  -\-  q. 
The  point  (p,  q)  is  the  vertex  of  the  parabola.  If  a  is  positive, 
the  vertex  is  a  minimum  point,  if  negative,  a  maximum  point  of 
the  curve.     Let 

y  =  Sx^  -  12x  +  19. 
y  =  3ix-  2)2  H-    7. 

The  last  equation  shows  at  once  that  the  minimum  value  of  the 
function  is  7  and  that  it  occurs  when  x  =  2. 

Exercises 

Find  the  maximum  or  minimum  values  of  the  following: 

1.  y  =  3x2  _  2x  +  1. 

2.  y  =  3x  -  2x2  +  1. 

3.  y  =  3x2  _^  7a., 

4.  If  a  body  is  thrown  vertically  upward  with  an  initial  velocity  of 
a  feet  per  second,  its  height  h  in  feet  at  the  end  of  t  seconds  is  given  by 

h  =  at  -  16.1<2. 

To  what  height  will  the  body  rise  if  thrown  with  an  initial  velocity  of 
32.2  feet  per  second?     When  will  it  reach  this  height? 

90.  The  Function  a  cos  x  +  b  sin  x.  The  function  a  cos  x  + 
6  sin  X  is  of  frequent  occurrence.      If  it  is  put  in  the  form  of 

165 


166 


CALCULUS 


^90 


the  product  of  a  constant  by  the  cosine  of  a  variable  angle,  the 
maximum  and  minimum  values  can  be  found  at  once.     Thus 


cos  X  +  —r= 


a  cos  X  -\-hsmx  =  ^aP'  +  h''' 

a  b 

Now, — /  and  — , 

'Va^  +  b^         Va^  +  b^ 

sine,  respectively,  of  an  angle  a.     For  if  P,  Fig. 
62,  be  the  point  (a,  6)  and  the  angle  POX  be  a, 


Va^  +  62 


sin  X 


[Va^  +  b^ 

may  be  regarded  as  the  cosine  and 


and 


Va^  +  b^* 
b 


Va^  +  b^  Fig.  62. 

Hence 

a  cos  X  -\-  b  sin  x  =  y/a^  -\-  b^  (cos  x  cos  a  +  sin  x  sin  a), 


or 


a  cos  X  -r  b  sinx  =  -x/a^  -j-  b^  cos  (x  —  a). 


(1) 


The  quadrant  in  which  ce  lies  will  be  determined  by  the  signs  of 
a  and  b. 

a  is  in  the  first  quadrant  if  a  is  positive  and  6  is  positive. 
a  is  in  the  second  quadrant  if  a  is  negative  and  b  is  positive. 
a  is  in  the  third  quadrant  if  a  is  negative  and  b  is  negative. 
a  is  in  the  fourth  quadrant  if  a  is  positive  and  b  is  negative. 

In  polar  coordinates  equation  (1)  shows  that  the  function 
a  cos  X  +  bsinx  is  represented  by  a  circle  passing  through  the 
pole,  of  diameter  -y/a^  +  b^,  and  with  its  center  on  the  line  making 
an  angle  a  with  the  polar  axis. 

The  right-hand  side  of  equation  (1)  shows  that  the  function  is 
represented  graphically  in  rectangular  coordinates  by  a  cosine 
curve  of  amplitude  -y/a^  -f  b^.  Thus,  the  maximum  value  of 
a  cos  X  +  &  sin  a;  is  \/a^  +  b^  and  occurs  when  x  =  a.  The  mini- 
mum value  of  the  function  is  —  \/a'^  +  b^  and  occurs  when 
2  =  a  -|-  X. 

Two  examples  giving  rise  to  this  function  are  solved  below. 


§90]  MAXIMA  AND  MINIMA  167 

Illustration  1.  The  weight  W,  Fig.  63,  rests  upon  the  horizontal 
surface  AB.  P  is  the  force,  inclined  at  an  angle  6  with  the 
horizontal,  which  will  just  cause  the  weight  to  slide  over  the 
plane.  The  problem  is  to  find  the  angle  6  for  which  P  will  be  a 
minimum.     The  coefficient  of  friction  is  denoted  by  ^i. 

The  normal  pressure,  N,  between  the  weight  and  the  plane  is 
(]F  —  P  sin  0),  the  difference  between  W  and  the  vertical  com- 
ponent of  P.  The  force  of  friction,  F,  is  then  fi  {W  —  P  sind). 
The  horizontal  component  of  P  equals  F.     Hence 

niW  -  Psind)  =  Pcosd, 
or 

^  =  cos0+Atsine.  (2) 

uW 
Since  ix  and  W  are  constants,-p-  is  a  maximum  when  and  only 

when  P  is  a  minimum.     Hence  to  find  the  minimum  value  of  P 

uW 
we  may  find  the  maximum  value  of  -p-  and  multiply  its  reciprocal 

hy  fiW 


W 


p 

B 


A < — i:i^itiw^m i B 


N 

Fig.  63.  Fig.  64. 


From  (2), 


^  =  Vl  +  M^' cos  (0  -  a), 


where  a  is  an  angle  in  the  first  quadrant  whose  tangent  is  /x, 
the  coefficient  of  friction.     Therefore,  when  6  is  acute  and  equal 

to  tan~^  /x,  JP  is  a  minimum  and  equal  to      , 

Vl  +  /X'' 

Illustration  2.  A  weight,  W,  Fig.  64,  rests  upon  the  inclined 
plane  AB.  Find  d  so  that  P,  the  force  which  will  just  cause  W 
to  move  up  the  plane,  shall  be  a  minimum. 

The  normal  pressure,  N,  between  the  weight  and  the  plane  AB 


168  CALCULUS  I  §91 

is  equal  to  Tf  cos  (3  -  P  sin  6.  Then  F,  the  force  of  friction,  is 
equal  to 

H  iW  cos  iS  -  P  sin  6), 

where  /t  is  the  coefficient  of  friction.     We  have  then 

F  =  m(^  cos  |8  -Psin0). 

Since  the  force  of  friction  must  balance  the  components  of  P 
and  W  parallel  to  the  plane  AB,  we  have 

H{W  cos  i3  -  P  sin  d)  =  (P  cos  e  -W  sin  /3). 

Hence 

Win  cos  j8  +  sin  /3) 

— ^^- p ^-^  =  cos  0  +  /i  sm  6, 

or 

Tr(^cos^  +  sin/?)^^^^-^^^^^^_^^  (3^ 

where  a  is  the  acute  angle  whose  tangent  is  n.  Thus,  the  left- 
hand  side  of  (3)  is  a  maximum,  and  P  a  minimum,  when  d  is  acute 
and  equal  to  tan~^  ju. 

„,  .     .  .              ,        f  p  •    .K      Tr(MCos^  +  sini8) 
The  mmimum  value  of  P  is  then  . 

Exercise 

In  Fig.  64,  find  the  minimum  force,  P,  and  the  angle  between  its 
line  of  action  and  AB,  which  will  just  prevent  the  weight  W  from  slid- 
ing down  the  plane. 

91.  The  Function  mx  +  -y/a^  —  x^.  Frequently  problems  in 
maxima  and  minima  lead  to  functions  of  the  form  mx  ±  ■s/a'^  —  x^. 
The  curve  for 

y  =  mx  ±  y/a^  —  x"^  (1) 

can  be  obtained  by  shearing  the  circle  y  =  +  \/a^  —  x^  in  the 
line  y  =  mx.  Every  ordinate  of  the  circle  to  the  right  of  the  F-axis 
is  increased  (or  decreased  if  m  is  negative)  by  an  amount  propor- 
tional to  the  distance  from  the  F-axis.  To  the  left  of  the  F-axis 
the  ordinates  are  decreased  if  m  is  positive  and  increased  if  m  is 
negative. 

The  maximum  value  of  y  is  easily  found  by  placing 

X  =  a  cos  t.  (2) 


§93]  MAXIMA  AND  MINIMA  169 

Then  from  (1) 

y  =  a(m  cos  t  +  sin  t)  (3) 

=  aVl  +  m^  cos  (t  —  a),  (4) 

where  a  =  tan"^  —     The  maximum  value  of  y  occurs  when 


X  =  a  cos  a  = 


vTT 


92.  Maxima  and  Minima  by  Limits  of  Curve.  In  case  f(x,  y)  =  0 
is  of  the  second  degree  in  x  and  y,  and  in  a  few  other  cases,  the 
maximum  and  minimum  values  of  y  can  be  found  by  determining 
when  X  changes  from  real  to  complex  values. 

The  method  will  be  illustrated  by  an  example. 
Let 

^  g^  +  e 

^  ~  2x  +  l" 
Then 

x  =  y  ±  V{y  +  S){y-2).  (1) 

From  equation  (1)  it  is  seen  that  for  values  of  y  greater  than  2 
or  less  than  —  3,  x  has  two  distinct  real  values.  When  y  =  2  or 
—  3,  X  has  two  equal  real  values.  When  —  3  <  ?/  <  2,  a;  is  imagi- 
nary. This  shows  that  the  line  y  =  c  meets  the  curve  in  two 
distinct  points  if  it  is  more  than  two  units  above  or  more  than 
three  units  below  the  X-axis;  that  it  is  tangent  to  the  curve 
when  two  units  above  and  when  three  units  below  the  X-axis,  and 
that  it  does  not  cut  the  curve  when  it  falls  within  the  limits  two 
units  above  and  three  units  below  the  X-axis.  Hence  the  func- 
tion has  a  minimum  value  2  and  a  maximum  value  —3. 

Exercises 

1.  Find  the  luaximum  and  minimum  values  of 

x"  -  2x  +  19 
2z  +5 

2.  Find  the  maximum  rectangle  which  can  be  inscribed  in  a  circle 
of  radius  10. 

93.  Maxima  and  Minima  Determined  by  the  Derivative.  The 
first  derivative  has  been  used  to  determine  the  value  of  the  argu- 


170  CALCULUS  [§93 

ment  corresponding  to  maxima  and  minima  of  functions.  Im- 
mediately to  the  left  of  a  maximum  point  the  function  is  increas- 
ing with  X  and  consequently  the  first  derivative  is  positive.  On 
the  other  hand,  immediately  to  the  right  of  such  a  point  the  func- 
tion is  decreasing  as  x  increases  and  the  first  derivative  is  negative. 
Similarly  it  follows  that  the  first  derivative  is  negative  im- 
mediately to  the  left  and  positive  immediately  to  the  right  of  a 
minimum  point.  In  both  cases  the  first  derivative  changes  sign 
as  the  independent  variable  passes  through  the  value  for  which 
the  function  has  a  maximum  or  a  minimum  value.  This  change 
of  sign  may  take  place  in  a  number  of  ways. 
Illustration  1.     Thus,  in  the  case  of  the  function 

y  =  x^  -2x  +  7, 
the  derivative, 

I  =  2.  -  2  .  2(.  -  1), 

is  negative  to  the  left  and  positive  to  the  right  of  the  line  x  =  1, 

dy 
When  a;  =  1,  ^  =  0  and  the  curve  has  a  horizontal  tangent.     In 

the  vicinity  of  this  point  the  curve  has  the 
shape  shown  in  Fig.  65. 

At  first  thought  it  might  appear  that  if  the 
first  derivative  is  negative  to  the  left  and 
positive  to  the  right  of  a  certain  point,  it 
certainly  must  become  zero  at  this  point. 
This  is,  however,  by  no  means  the  case,  as 
the  next  illustration  will  show. 

Illustration  2.  y  =  4  +  (x  —  1)^.  Although  the  minimum 
value  of  this  function  can  be  determined  at  once  by  noting  that 
it  represents  a  semi-cubical  parabola  with  its  vertex  at  (1,  4),  the 
problem  will  be  worked  by  the  method  of  the  calculus  for  illustra- 
tive purposes. 
The  derivative, 

dy^ 2 ^ 

^^     3(x-l)*' 

is  negative  when  x  <  1  and  positive  when  x  >  1.     Hence  the 
function  is  decreasing  to  the  left  and  increasing  to  the  right  of 


§93] 


MAXIMA  AND  MINIMA 


171 


a;  =  1.  When  x  =  1,  ?/  =  4.  This  value  is  a  minimum  value  of 
the  function.  For  z  =  I  the  derivative  does  not  exist,  as  the 
denominator  becomes  zero.    Let  us  see  what  really  happens  in 

the  vicinity  of  x  =  1.    As  x  approaches  1  from  the  left,  j-  takes 

on  larger  and  larger  negative  values.  The  form  of  the  curve  to 
the  left  and  in  the  immediate  vicinity  of  the  point  (1,  4)  is  some- 
thing like  that  shown  in  Fig.  66.  The  line  x  =  1  is  tangent  to 
the  curve  at  this  point. 

As  X  approaches  1  from  the  right,  i.e.,  through  decreasing  values, 
the  derivative  becomes  larger  and  larger.  The  form  of  the  curve 
to  the  right  of  the  line  x  =  1  is  also  shown  in  Fig.  66.  The  line 
X  =  1  is  also  tangent  to  the  portion  of  the  curve  obtained  by  allow- 
ing X  to  approach  1  from  the  right. 


Fig.  C6. 


Fig.  67. 


It  is  now  apparent  that  the  first  derivative  may  change  sign 
without  passing  through  zero.  In  the  above  illustration  it  changes 
sign  by  becoming  infinite. 

The  first  derivative  may  change  sign  in  still  another  way  as 
illustrated  by  the  curve  of  Fig.  67.  Let  us  suppose  that  the  de- 
rivative approaches  —  1  as  x  approaches  a  from  the  left,  and  the 
value  +1  as  X  approaches  a  from  the  right.  The  function  has  a 
minimum  value  at  the  point  P,  for  the  derivative  changes  from 
minus  to  plus  as  x  increases  through  the  value  a  and  consequently 
the  function  is  decreasing  to  the  left  and  increasing  to  the  right 
of  X  =  a. 

The  essential  thing  at  a  minimum  point  is  that  the  derivative 
changes  sign  from  minus  to  plus,  and  at  a  maximum  point  that  it 
changes  sign  from  plus  to  minus. 


172  CALCULUS  [§95 

A  derivative  which  is  continuous  at  a  maximum  or  a  minimum 
point  changes  sign  by  passing  through  zero.  But  it  may  change 
sign  by  becoming  infinite,  as  the  second  illustration  shows,  or  by 
becoming  otherwise  discontinuous  as  explained  above.  This  last 
type  is  of  rare  occurrence  and  will  not  be  referred  to  again. 

Illiistration  3.    y  =  x^  -^  3.    The  derivative  of  this  function, 

-p  =  3x^,  is  positive  for  all  values  of  x  except  x  =  0,  when  it  is 

zero.  The  function  is  increasing  for  all  these  values  of  x.  At 
this  point,  (0,  3),  there  is  a  horizontal  tangent  but  the  function 
has  neither  a  maximum  nor  a  minimum  at  the  point,  for  it  in- 
creases up  to  the  value  3  for  x  =  0  and  then  continues  to  increase 
to  the  right.  This  illustration  brings  out  clearly  the  fact  that 
there  is  no  reason  for  assuming  that  a  function  has  a  maximum  or 
a  minimum  value  at  a  point  where  the  first  derivative  is  zero. 
What  kind  of  a  point  is  the  point  (0,  3)  ? 

94.  Second-Derivative  Test  for  Maxima  and  Minima.  In  the 
first  of  the  three  types  of  maximum  or  minimum  points  considered 
in  §93,  the  first  derivative  changes  continuously  from  positive  to 
negative  values  or  vice  versa.  For  a  maximum  point  of  this  type 
the  curve  is  concave  downward  and  the  second  derivative  is 
negative  at  such  a  point.  For  a  minimum  point  the  curve  is 
concave  upward  and  the  second  derivative  is  positive.  A  con- 
venient test  for  the  behavior  of  a  function  at  a  point  where  the 
first  derivative  is  zero  is  then,  to  substitute  the  abscissa  of  this 
point  in  the  expression  for  the  second  derivative.  If  the  second 
derivative  is  positive  the  point  is  a  minimum  point;  if  negative, 
a  maximum  point.  If  the  second  derivative  is  zero,  the  test 
fails.  This  test  also  fails  for  maximum  or  minimum  points  where 
the  first  derivative  is  discontinuous. 

Examine  the  curves 

y  =  x», 

y  =  X*, 


95.  Study  of  a  Function  by  Means  of  its  Derivatives.     The 

following  is  a  summary  of  the  application  of  the  first  and  second 
derivatives  to  tracing  the  curve  representing  a  function: 


§95] 


MAXIMA  AND  MINIMA 


173 


1.  The  function  is  increasing  if  the  first  derivative  is  positive, 
and  decreasing  if  it  is  negative. 

2.  To  find  maximum  and  minimum  points  find  the  values 
of  X  for  which  the  first  derivative  becomes  zero  or  infinite.  If 
the  derivative  changes  sign  at  any  of  these  points,  the  correspond- 
ing point  is  a  maximum  or  minimum  point  according  as  the 
change  is  from  plus  to  minus  or  vice  versa. 

Points  at  which  the  first  derivative  is  equal  to  zero  can  also  bo 
tested  by  substituting  the  abscissa  of  the  points  in  the  second 
derivative.  If  the  second  derivative  is  positive,  the  point  is  a 
minimum  point,  if  negative,  a  maximum  point. 

3.  Points  of  inflection  are  found  by  determining  where  the 
second  derivative  changes  sign.  As  in  the  case  of  the  first 
derivative,  the  change  in  sign  can  take  place  through  zero  or 
infinity.  If  the  change  is  from  positive  to  negative  values  the 
curve  changes  from  being  concave  upward  to  being  concave 
downward. 

The  abscissas  of  the  points  at  which  the  first  and  the  second 
derivatives  become  zero  or  infinite  we  shall  call  the  critical 
values.  These  values  and  these  alone  need  be  tested  in  study- 
ing the  behavior  of  an  ordinary  curve.  The  investigation  of  a 
curve  by  means  of  its  derivatives  can  be  put  in  the  tabulated 
form  shown  in  the  following  illustrative  examples: 


1.  y 


(See  Fig.  33.) 

dy 

dx 

d^ 

dx^ 


—  X. 


dy 
dx 


=  0  when  x  =  0. 


-1-^  =  0  when  x 


X 

dhj 
dx^ 

dy 
dx 

y 

X  <0 
X  >  0 

<0 
>0 

>0 

>o 

Concave  downward,  increasing 
Concave  upward,  increasing 

174 


CALCULUS 


^96 


Here  (0,  0)  is  a  point  of  inflection.     There  is  neither  a  maxi- 
mum nor  a  minimum  point. 
2.  1/  =  ix3  -  x^  +  Ix  +  2.     (See  Fig.  34.) 

g  =  ix^  -  2x  +  I 


=  X 


i(:r  -  Dix  -  3). 
2. 


dv 

-^  =  0  when  x 

ax 


=  0  when  x 


1,3. 
2. 


X 

dy 
dx 

y 

X  <  2 
X  >  2 

<o 

>  0 

Decreasing 
Increasing 

Concave  downward 
Concave  upward 

X  <  1 

1  <x  <  3 

X  >  3 

>  0 
<  0 

>  0 

Increasing 
Decreasing 
Increasing 

1 
2 
3 
0 

8 
3 

i 

2 
2 

(2,  I)  Point  of  inflection. 
(1,  f)  Maximum  point. 
(3,  2)  Minimum  point. 

Apply  second  derivative  test  for  x  =  1  and  x  =  3. 

96.  Applications  of  Maxima  and  Minima.  In  solving  problems 
involving  maxima  and  minima  the  first  step  is  to  set  up  from  the 
conditions  of  the  problem  the  function  whose  maximum  or  mini- 
mum value  is  sought.  Frequently  the  function  will  be  expressed, 
at  first,  in  terms  of  two  or  more  variables.  Usually,  however, 
there  is  a  relation  between  these  variables,  and  the  function  can  be 


§96] 


MAXIMA  AND  MINIMA 


175 


expressed  in  terms  of  a  single  variable.     After  this  has  been  done 
the  maximum  or  minimum  values  can  be  found. 


Exercises 

1.  Equal  squares  are  cut  from  the  comers  of  a  rectangular  piece  of 
tin  30  by  20  inches.  The  rectangular  projections  are  then  turned  up 
forming  the  sides  of  an  open  box.  Find  the  size  of  the  squares  cut  out 
if  the  volume  of  the  box  is  a  maximum. 

2.  A  man  who  is  in  a  boat  3  miles  from  the  nearest  point,  A,  of  a 
straight  shore  wishes  to  reach,  in  the  shortest  possible  time,  a  point 
B  on  the  shore  which  is  6  miles  from  A.  Find  the  point  of  the  shore 
toward  which  he  should  row,  if  he  can  row  at  the  rate  of  3  miles  per 
hour  and  walk  at  the  rate  of  5  miles  per  hour. 

3.  The  horizontal  component  of  the  tension  in  the  guy  wire  BC, 
Fig.  68,  is  to  balance  the  horizontal  pull  P.  If  the  strength  of  the  wire 
varies  as  its  cross  section,  and  if  its  cost  varies  as  its  weight,  find  the 
angle  0  such  that  the  cost  of  the  guy  wire  shall  be  a  minimum. 

B 


Fig.  68. 


Fig.  69. 


4.  Find  the  length  of  the  shortest  beam  that  can  be  used  to  brace 
a  wall  if  the  beam  passes  over  a  second  wall  6  feet  high  and  8  feet  from 
the  first. 

5.  A  steel  girder  30  feet  long  is  moved  on  rollers  along  a  passageway 
10  feet  wide,  and  through  the  door  AB,  Fig.  69,  at  the  end  of  the  pas- 
sageway. Neglecting  the  width  of  the  girder,  how  wide  must  the 
door  be  in  order  to  allow  the  girder  to  pass  through? 

6.  A  sign  10  feet  high  is  fastened  to  the  side  of  a  building  so  that  the 
lower  edge  is  25  feet  from  the  ground.  How  far  from  the  building 
should  an  observer  on  the  ground  stand  in  order  that  he  may  see  the 
sign  to  the  best  advantage,  i.e.,  in  order  that  the  angle  at  his  eye  sub- 
tended by  the  sign  may  be  the  greatest  possible?  The  observer's  eye 
is  5 1  feet  from  the  ground. 

7.  A  man  in  a  launch  is  m  miles  from  the  nearest  point  A  of  a 
straight  shore.  Toward  what  point  on  the  shore  should  he  head  his 
boat  in  order  to  reach,  in  the  shortest  possible  time,  an  inland  point 


176 


CALCULUS 


[§96 


whose  distance  from  the  nearest  point  B  of  the  shore  is  n  miles?     The 
man  can  run  the  boat  Vi  miles  per  hour  and  can  walk  Vj  miles  per  hour. 
The  distance  AB  is  p  miles. 
Ans.  Toward  a  point  such  that 

sin  61       sin  dt 


Vl 


V2 


where  0i  and  62  are  the  angles  made  by  the  paths  of  the  man  with  the 
normal  to  the  beach.  It  will  be  noticed  that  the  path  taken  by  the 
man  is  similar  to  that  followed  by  a  ray  of  light  in  passing  from  one 
medium  to  another  with  a  different  index  of  refraction. 


-.25 


.5  1  1.5 

Fig.  70. 

8.  A  man  in  a  launch  is  m  miles  from  the  nearest  point  A  of  a 
straight  shore.  He  wishes  to  touch  shore  and  reach,  in  the  shortest 
possible  time,  a  second  point  on  the  lake  whose  distance  from  the 
nearest  point  B  on  shore  is  n  miles.  In  what  direction  must  he  head 
his  boat  if  the  distance  AB  is  p  miles? 

The  path  taken  by  the  man  is  similar  to  the  path  of  a  ray  of  light 
reflected  by  a  plane  surface. 

9.  It  is  desired  to  make  a  gutter,  whose  cross  section  shall  be  a 


MAXIMA  AND  MINIMA 


177 


segment  of  a  circle,  by  bending  a  strip  of  tin  of  width  a.     Find  the 
radius  of  the  cross  section  of  maximum  carrying  capacity. 

10.  A  sector  is  cut  from  a  circular  piece  of  tin.  The  cut  edges  of 
the  remaining  portion  of  the  sheet  are  then  brought  together  to  form  a 
cone.  Find  the  angle  of  the  sector  to  be  cut  out  in  order  that  the 
volume  of  the  cone  shall  be  a  maximum. 

11.  The  stiffness  of  a  rectangular  beam  varies  as  its  breadth  and  as 
the  cube  of  its  depth.  Find  the  dimensions  of  the  stiffest  beam  which 
can  be  cut  from  a  circular  log  12  inches  in  diameter. 

12.  The  strength  of  a  rectangular  beam  varies  as  its  breadth  and  as 
the  square  of  its  depth.  Find  the  dimensions  of  the  strongest  beam 
which  can  be  cut  from  a  circular  log  12  inches  in  diameter. 

1.5 


L26 


.75 
(4) 


.25 


k^^ 

3^(2) 

(1)  2/  =  e"* +0.56-^"* 

(2)  y=e- 

(3)  2/  =  e  ^-o.ie 

(6)  2/ =  e-* -1.56  ^ 

>A(3) 

p^ 

\, 

|(6) 

\ 

s^ 

— - 

1  1.5 

Fig.  71. 


2.6 


13.  Consider  the  sum 

y  =  ox"  +  bx*" 
for  positive  values  of  x  only.  First,  if  n  and  r  are  of  like  sign,  show 
that:  (1)  a  maximum  or  a  minimum  value  exists  if  a  and  b  are  of 
unlike  sign;  (2)  neither  a  maximum  nor  a  minimum  value  exists  if  a 
and  b  are  of  like  sign.  Second,  discuss  the  same  cases  if  n  and  r  have 
opposite  signs. 

14.  Determine  the  exact  values  of  the  maxima  shown  in  Figs.  70 
and  71. 

Hint.     Consider  first  the  general  case 


12 


y 


e~'  —  ae 


CHAPTER  XI 

POLAR  COORDINATES 

97.  Direction  of  Curve  in  Polar  Coordinates.  Let  BPQ, 
Fig.  72,  be  a  curve  referred  to  0  as  pole  and  OA  as  polar  axis. 
Let  P  be  any  point  of  the  curve  and  let  PT  be  a  tangent  to  the 
curve  at  this  point.  Let  PS  be  the  radius  vector  of  the  point  P, 
produced. 
A  point  describing  the  curve,  when  at-P,  moves  in  the  direction 
PT.  This  direction  is  given  by  the 
angle  ^  through  which  the  radius 
vector  produced  must  rotate  in  a  posi- 
tive direction  about  P,  in  order  to  be- 
come coincident  with  the  tangent  line. 
An  expression  for  tan^  will  now  be 
found.  Let  Q,  Fig.  72,  be  a  second 
point  of  the  curve.  PR  is  perpendicu- 
lar to  OQ,  and  PM  is  a  circular  arc  with  0  as  center  and  radius 
OP  =  p. 

tan^=  iJ^otani2QP=   iJS'o  ^.  (D 

The  infinitesimals  PR  and  RQ  can  be  replaced  by  PM  and  MQ, 
respectively,  if  (see  §60) 

lim  PR 

A0=O  PM 


Fig.  72. 


1 


(2) 


and 


=  1. 


lim   RQ 

Equation  (2)  is  true  by  equation  (3),  §56 
(3)  follows: 

lim   RQ^ 

Afl-.0  MQ 


(3) 
The  proof  of  equation 

lim  RM  +  MQ 


Ae=o        MQ 

Urn  p(l  -cos  Ag)  +  Ap 

A9=0  ^p 

,    ,    lim  p(l-cosA0)  Ae 
178 


Ad 


Ap 


§97] 
Hence 

since 


POLAR  COORDINATES 


179 


lim  RQ 


=  1. 


Ae=o  MQ 
lim  1  —  cos  Ad 

A9=0  ^0 

From  (1),  (2),  and  (3)  it  follows  that 


=  0. 


,       ,        lim  PM 


Hence 


lim  pAg 


tan^ 


dp 


tan^  =  ^- 
dd 


(4) 


This  formula*  can  be  easily  remembered  if  the  sides  of  the  tri- 
angular figure  MQP,  Fig.  72,  are  thought  of  as  straight  lines,  and 


d  sin  0 
'  This  formula  enables  us  to  give  another  proof  for  — -yz — .    In  polar  codrdinates 

p  =  sin  8  represents  a  circle,  Fig.  73.     By  geometry,  x//  =  e.     Then 

p  sin  0 

tan^  =  tanfl=^   -   ^. 

de 


dp 

de  "■ 

d  sin  0 

de 


de 


cos  8, 


180  CALCULUS  [§97 

the  angle  MQP  as  equal  to  \p.     Then  the  tangent  of  rp  would  be 


de  p_ 
dp 
dd 


dp      dp 


dv 
Formula  (4)  corresponds  io  -r  =  tan  t  in  rectangular  coordi- 
nates. 

Illustration  L     If  p  =  e"*, 

dd  =  «^ 

and  tan  \I/  =  -,  a,  constant. 

^       a' 

Illustration  2.  Find  the  equation  of  the  family  of  curves  for 
which  the  angle  between  the  radius  vector  produced  and  the 
tangent  line  is  a  constant. 

tan  yp  =  k. 


or 


Integrating 


or 


dp 
dd 

k, 

dp 

; 

dd  _ 
P 

1 

k 

dp 
P 

>• 

IP  = 

1-- 

P  = 

«:— 

= 

e    e*, 

P  = 

Kek, 

where  K  is  an  arbitrary  constant. 


^98]  POLAR  COORDINATES  181 


Exercises 

Find  tan  \p  for  each  of  the  following  curves : 

1            '^. 

4.  p  =  a(l  —  cos  e). 
5                o 

2.  p  =  ad. 

''      1  -  cos  » 

3.  p  =  e"'. 

6.  p  =  o  cos  (0  —  a). 

98.  Differential  of  Arc:    Polar  Coordinates.     We.  shall  now 

ds 
find  an  expression  for  -v^  in  polar  coordinates.     From  Fig.  72, 


dd 
(chord  PQ)  2  =  (PRy  +  (RQ^. 


From  which 


lira    /chord  PQV        lira   /PRy   ,     lira   /RQV 

Replacing  chord  PQ  by  arc  PQ  =  As,  PR  by  PM  =  pA0,  and 
RQ  by  MQ  =  Ap, 


(1) 
(2) 

(3) 

It  corresponds  to  (ds)^  =  (da-)^  +  (%)'' in  rectangular  coordinates. 
It  can  be  remembered  easily  by  the  help  of  the  triangle  MQP, 
Fig.  72. 

The  length  of  the  curve  can  be  expressed  as  a  definite  integral. 
Thus:  (See  Fig.  74) 

lim    -^    _,^ 

s  =   A«  =  0    2Lf    P^ 


lira  / 

A9  =  0  1 

Y       lim  fpAey  ,    1 

Therefore 

[de)  =  ^  +  [de) 

and 

dS   =   Jp2+(^P)W 

This  formula  can 

be  written 

(ds)2  =  p2(d0)2  +  (dp)^ 

182  CALCULUS  §98] 

=  a/™o    %  ViPRV  +  {RQy 


B 
=  /3 


-J'i-0  D'VQV  (ff)'- 


=  a 

0  =  fi 


Illustration.  Find  the  entire  length  of  the  curve  p  =  a(l  —  cos  0). 
This  curve  is  symmetrical  with  respect  to  the  polar  axis.  The 
length  of  the  upper  half  will  be  found  and  multiplied  by  2. 

dp 


=    j    Va\i  -  cos  ey  +  a2  sin2  d  dd 


=^^de 


a 

sin  -H"  rf0 


=  —  4a  cos  —     =  4a. 
^  lo 
s  =  8a. 

Exercises 

1.  Find  the  entire  length  of  the  curve  p  =  2a  sin  9. 

2.  Find  the  entire  length  of  the  curve  p  =  a(l  —  sin  6). 

a 

3.  Find  the  entire  length  of  the  curve  p  =  a  sin'  -^• 

o 

4.  Find  the  length  of  p  =  e°*  between  the  points  corresponding  to 
6=0  and  0  =  v.     Also  between  the  points  corresponding  to  0  =  0 

and  e  =Z- 


§99] 


POLAR  COORDINATES 


183 


5.  Prove  formula  (3)  directly  from 


and 


X  =  p  cos  d, 
y  =  p  sin  e, 

ds  =  Vidxy  +  (dj/)2. 


99.  Area :  Polar  Coordinates.  Find  the  area  bounded  by  the 
curve  p  =  fid)  and  the  radii  vectores  6 
=  a  and  6  =  ^.  We  seek  the  area 
BOC,  Fig.  74.  Draw  radii  vectores 
dividing  the  angle  BOC  into  n  equal 
parts  Ad.  Let  POQ  be  a  tj'pical  one 
of  the  n  portions  into  which  the  area 
is  divided  by  these  radii.  The  angle 
POQ  is  Ad.  The  line  OP  makes  an 
angle  6  with  the  initial  line  OA,  and  ^^°-  '^' 

its  length  is  p  =  f(d).     Denote  the  area  of  BOC  by  A. 

lim 


o7 


^  =nT^XPOQ^ 


(1) 


Replace^  POQ  by  the  circular  sector  POR  whose  area  is  ^p^'A^. 

Then 

0  ^  s 


or 


'I 


P^dd, 


A  =  h\  [mv dd. 


Exercises 


(3) 


1.  Find  the  area  bounded  by  the  curve  p  =  2o  sin  e. 

2.  Find  the  area  bounded  by  the  cardioid  p  =  2a(l  —  cos  6). 

,          1  +  cos  29 
Hint.     cos*0  = ^ 


iLetA^   =  OPQ  (Fig.  74). 


PR  and  QS  are  arcs  of  circles.     Then 
OPR  <  A^  <  OSQ, 


Jp«Ae  <AA<  i(p  +  Ap)'AJ. 


184  CALCULUS  [§99 

3.  Find  the  area  bounded  by  p  =  2a(l  +  sin  e). 

4.  Find  the  area  bounded  by  one  loop  of  p  =  10  cos  2d. 

5.  Find  the  area  bounded  by  one  loop  of  p  =  10  sin  26, 

6.  Find  the  area  bounded  by  one  loop  of  p  =  a  cos  Sd. 

7.  Find  the  area  bounded  by  one  loop  of  p^  =  10  cos  29. 

8.  Find  the  area  swept  out   by  the   radius   vector  of   the  curve 
p  =  ad,  as  0  varies  from  0  to  2ir. 

9.  Find  the  area  bounded  by  the  radii  vectores  0  =  2,  ^  =  ir,  and 

the  curve  p  =  -r^- 

10.  Find  the  area  bounded  by  the  radii  vectores  0  =  0,  6  =  ^  and 
the  curve  p  =  5^^. 


CHAPTER  XII 

INTEGRATION 

100.  Formulas.  In  Chapters  III,  VI  and  VII  the  following 
formulas  of  integration,  with  the  exception  of  (19),  have  been  used. 
They  are  collected  here  foj  reference,  and  should  be  memorized  by 
the  student. 

1.    fw  du  =  — 7-r  w»+i  +  C,  if  w  7^  -  1. 
J  n  +  1 


3.  J  e»  rfw  =  e»  +  C. 

4.  I  a"  du  =  , a»  +  C. 

J  log, a 

5.  j  sin  udu  =  —  cos  u  -\-  C. 

6.  J  cos  w  dw  =  sin  w  +  C. 

7.  I  sec'^  udu  =  tan  u  +  C. 

8.  I  csc'^  udu  =  —  cot  w  +  C. 

9.  I  sec  u  tan  udu  =  sec  m  +  C 

10.  I  esc  M  cot  udu  =  —  esc  w  +  C 

11.  J  tan  udu  =  log  sec  w  +  C. 

12.  J  cot  udu  =  log  sin  w  +  C 

13.  J  sec  udu  =  log  (sec  w  +  tan  u)  +  C 

14.  { CSC  udu  =  —  log  (esc  w  +  cot  u)  +  C. 

15.  I  —r =  sm-i  -  +  C. 


185 


186 

CALCULUS 

16. 

du 
a^  +  u^ 

=  -  tan-i  -  +  C. 
a            a 

[§100 


17.  I .-  =  -  sec-i  -  +  C. 

J  u\/u'^  —  a2      «  « 

18.  I     ,J^ =  log  (m  +  Vu^  ±  a^)  +  C. 


r  du   ^  1^ 

J  u^-a^~  2a 


19.    Iri^.-^loK^  +  C.ifX- 

=  K-  log  — j \-  C,itu  <  a. 

2a     ^  a  +  u 

Formula  (19)  is  proved  as  follows: 


.^^lU Ll 

u2  —  qZ       2a  \u  —  a       u-\-  aj 

f  du  ^  i_  rr_i 1  1 

J  u^  -  a^       2aJ  [u-a       w  +  a  J   " 

_  1  r  du      1  r  du 

~  2a  J  u  —  a       2a  J  u-\-  a 

=  2"  log  {u  -  a)  -  2^  log  (w  +  a)  -f  C. 

1  ,      u-a   ^ 

=  ^  log  — , h  C. 

2a     ^  u  +  a 

This  formula  leads  to  the  logarithm  of  a  negative  number  if  m  <  a. 
To  obtain  a  formula  for  this  case  write 

^_  =  ir__i Li. 

u^  —  a^       2a  L      a  —  u       a-\-uJ 
Then 


/ 


du  1   ,       a  —  M    .    _ 

=  o:;  log  :,-n-7:  +  C- 


■u^  —  a^       2a     ^  a  +  w 
Exercises 


r   xdx  r 

*  J  Vie  -  z2*  ^*  J  V^ 

^*  J  Vie  -  X*"  *'  J  ^'  + 16 


2-16 

dx 


§100] 


INTEGRATION 


187 


r  dx 

•  J16-X 

I 


16 


xdx 
x^  -  16* 

dx 


xy/x^  -  16 

9.    fcot  1i  dt. 

r{x+  a)dx^ 
J    ^'  +  2ax 

11.  fix^  -  16)^  X  dx. 

12.  fsin  (2x  -  3)  dx. 

13.  fsec^  (5a  +  2)  da. 

25.  I     e2  +e~2   dx. 

26.  la^''  dx. 

27.  r(2x  +  4)''-"  dx 

^^'  J  7x*  +  ll 
30.    I 


4.  Jsec  (2(?  +  4)  tan  (20  +  4)  de. 

5.  fcsc^  (3  -  2</.)  d<^. 

7.    fe^^'^^ainede. 

C    31^  dt 
J  30<'+  13' 

9.  JiVc  -  Vxy  dx. 

20.  J  \/3  +  4x  dx. 

/I 
e^dy. 

23.    fe*'*"  (2^  +2)  sec2(2x  +  3)  dx. 


dx.     Divide  numerator  by  denominator. 
31.    le^'dx. 


dx 


/dx 
4x2  _ 

/ 

J  Vie  -  9x2" 

35.    ftan  (3a  +  4)  da. 


33 


34 


9 

cos  X  dx 
4  +  3  sin  X 

dx 


cos"  (3x  -  2) 
36.    Htan  e  +  cot  ey  dd  =  tan  e  -  cotB  +  C. 

39.  Jcos  (3<  -  4)  dt. 

„  40  r  ^*- 

38.  J  cot  (5<  -  8)  dt.  J  ^  -  *2/' 


/■ 


37.     I  (sin  -z  —  cos  50)  dd. 


188  CALCULUS  [§100 


dz  61.    fcos^  (3x  -  2)  sin  (3x  -  2)  dx. 

62.    I  sec*  (9  -  7x)  dx. 


63.    I  e      3  sec^^dx. 


•{.  63. 

43.  J  tan  (2x  -  5)  dx.  J 

44.  Jsec  (2y  +  4)  dt/.  64.  J  tan*  x  sec*  a:  dx. 
46.    fcsc  (2t/  —  7)  dy.  65.  J  cos*  3x  sin  3a;  dz. 

46.  J  cot  (3<  +  11)  d/.  66.    ftan^  5x  sec*  5xdx. 

47.  I  sec*  (^  —  5]dx.  67.    I  sec*  x  tan  xdx. 

48.  jTcos  (3  -  2x)  dx.  «8.  J  csc^  x  cot  x  dx. 

Aa     r      2x  +  5       ^  69.   fx  tan  (2x*  -  5)  dx. 
**•     I  x*  +  5x  +  4r^-  % 

/  _       I  sin  X  cos  X  dx 

62 


/<^^  72.    (e*   —^dx. 

V9<*  -  4  J        * 

/<^^  73.  fiJ  -  x»)» dx. 
V4-9/*- 

J  <\/9<*  -  4  "^        

/tdt  75.  ft^Qt^  -  17 dt. 

V9i^^4:  r   ^dx 

r  tdt  ^^J 

j9t*-4  n 

.    I  sec  5x  dx.  '   j  i 


V9  -X* 
sin  5x  dx 


57.    I  sec  5x  dx.  '  /  ^  ^*^^  ^^  "I"  ^^ 

68.  fsin  (cot  +  a)  dt.  78.  fe^'''  ^^  cos  3x  dx. 

69.  Jcos*  4x  sin  4x  dx.  79.  fe^'  +  <^  +  '^{x  +  3)  dx. 
60.   j  sin*  (x  +3)  cos  (x  +  3)  dx.  80.  |  cos  5x  sin  3x  dx. 


INTEGRATION 

83.  Jsin  3t  cos  4t  dt. 


84.    I  sin«5«d<. 


§100]  INTEGRATION  189 

81.  I  cos  3x  cos  5x  dx. 

82.  I  sin  7x  sin  4x  dx. 

r 

85.  I  sin  mt  cos  n<  di,  where  m  and  n  are   integers.     What  is  the 
value  if  m  =  n? 

86.  I  cos  (3w<  +  a)  sin  (3w<  +  a)dt. 


'I? 


88 


89 


90 


2x  +  3   , 

x^  +  3x^+7 
x*  +  9 


•J  3^^^^-     ■ 

/3x-f-2    , 

91.  fsin*  5fl  cos  56  dO. 

92.  I  sin^  X  dx. 

93.  J  xVie  -  x2  dx. 

94.  I  sec2  (l  +  2J  dx. 

95.jy^^dt. 

96.  I  sec  2  4x  tan  4x  dx. 

97.  I  \  dx. 

109.  fsec  {3<l>  -  2)  tan  (30  -  2)  d,/.. 

110.  rtanH2x  -  1)  sec*  C2x  -  1)  dx. 


dx 


V5  -  7x2 
dx 


VSx^  -  5 

00.  I  sin  4x  cos  6x  dx. 

01.  f(V^-V^)'dx. 
'2x  +  3 


2x  +  7 
dx 


dx. 


V3x  +2 

04.  I  sin*  2x  cos  2x  dx. 

05.  I  Vsinx  cos  x  dx. 

06.  fe-^dt. 

07.  r(2x  -  5)^dx. 


08. 


/ 


sec  odx. 


190  CALCULUS  [§101 

"^-  jl  +  tan3x^^-  "8-j3^M^4 

112.    fv2^^3idx.  119.     [ —y^ 

•^  J  yVsy 


dy 


113.    ftan  (5  -  2x)  dx.  f*      vdv 

J  120     I  —^=^ 


114 
116 


f    .i.  J  Vs.' -7 

J  5  -  3x2  ]L21.    I  sec^  0  tan  e  dd. 

•  J  ¥~^7  "2-  J  VlP^' 

116.    I^^dx.  123.    |sm'|cos|dx. 

/x  +  4  C  X 

— ; ?;  <ix.  124.     I  cos  2xsin  o<ia;. 

x^  -  9  J  2 

101.  Integration  of  Expressions  Containing  ax^  +  bx  +  c,  by 
Completing  the  Square. 
Illustration  1. 

/dx  ^  r         dx 1_  J  X  +  2 

X2  +  4X  +  9      J(x  +  2)2+5       VS  VS 

Illustration  2. 

/rfx  r  dx  C  dx 

V3  +  4x  -"x^  ~  J  VS  -  (x2  -  4x)  ~  J  \/3  +  4-(x-2)2 
dx  .        X  —  2 


■J: 


\/7  -  (x  -  2)2  \/7 

Illustration  3. 


+  C. 


r      dx      _  1  r 

J7x2  +  3x  +  ll     IJ 


x^  +  ^x  +  tEit  +  V  -  ~lU 
dx 


1      14     ^     _,x  +  -A-  ,   ^ 

s  — ;==  tan  *    .> .  +  C 

'  V^99  V^99^ 

14 


2 


tan 


-.  lii+3  +  c. 


299  \/299 


§101]  INTEGRATION  191 

Illustration  4. 

dx 


r      dx i_  r      dx 1^  r 

J  VQ+2x-3x^~ V3J  V2+lx-x^~VSJ  V2-{x^-l^ 

1     C  dx  1      .   _.  x-l 

=  — 7=  I  — ■—  =  — -7=  sin  ■*  — 7=  +  C 

\/3j  VY  -ix-\r       \/3  VY 


— ;-  Sin^  ;p:r—  +  C. 

V3  \/l9 


Exercises 


J  x^  +  6x  +25  J 


^-  J  x2  -  6x  +  5  J  5x^  - 

4.    f-^^ 9.    f 

J  A/2a;2  +  2x  -  3  J 


V2  +  3t  -  2t2 
dx 


8x  +  l 
dt 


Vl  +2t  +  2t2 


2x2  _|.  5x  -  3 


,„      r      2x-5        ^  f2x+6-ll  _, 

1°-  Jx^  +  6x+25^^  =J^2+6x+25'^^ 

/(2x  +  6)dx     _        r___^5___ 
x2  +  6x  +  25  J  x2  +  6x  +  25* 

J  V2x2^+  2x  -  3  J  VS  -  4x  -  x2 

/4x  +  1 1  C 

^^-p2^-dx.  14.  J- 

16  r     ^^  18.  f     ^^ 

'  J    xV3l^  6x  +  5x2  J  V4J/2  +  122/  -  7 

17.    f-^-='  19.    fg^- 


dx  1 

15.     I  ^ Substitute  x  =  -• 

xv/2x2  +  3x  -  2  ^ 


(Zx 

24x  -9 


192  CALCULUS  [§102 

20.    I  23.    I  .  ^=' 

J  V8  +  12x  -  4x»  J  x\/^+12x  -  7x2 

J  \/8  +  12x  -  4x2  J  a 


X  VSx^  +  12x  -  4 


/ 


22. 


16x2  -  24x  +  24 


102.  Integrals  Containing  Fractional  Powers  of  x  or  of  a  +  bx. 
Illustration  1. 


/: 


x^  —  x^ 
—  ax. 


3  +4 
Let  X  =  2*.     Then  dx  =  62^  dz,  and 

J    ^i  +  4  ^  j0^  +  4'  ^"  J^^  +  4  '''• 

The  integration  can  be  performed  after  dividing  the  numerator 
by  the  denominator  until  the  degree  of  the  remainder  is  less 
than  2.    After  integration  replace  z  by  x^. 
Illustration  2. 

\x  +  2)^  +  4 


/! 


dx. 


(x  +  2)^-3 
Let  X  +  2  =  z<.     Then 

r(x  +  2)U4  f(?!+-41^'rf. 

Divide  the  numerator  by  the  denominator.     The  integration  can 
readily  be  performed.     After  integration  replace  2  by  (x  +  2)*. 

In  general  if  fractional  powers  of  a  single  linear  expressior, 
a  +  bx,  occur  under  the  integral  sign,  let  a  -{-  bx  =  z",  where  n  is 
the  least  common  denominator  of  the  exponents  of  a  +  bx.  The 
linear  expression  a  +  bx  reduces  to  x  when  a  =  0  and  6=1. 
See,  for  example,  Illustration  1. 

Exercises 


§103]  INTEGRATION  193 


f. 
■■/' 


^„      ,      3x-  2       , 

10.     I         . dx. 

x\/2x + 3 


(a;i  +  a;3)  dx. 


12.    I .  dx. 

Sx\/2x  -  3 


14. 


J  1  +  (X  +  2)^ 


13.     I  v-^-T--^-   -rx^^^ 
1  +  (X  +  2) 
Vx  —  3  dx 


P 


x  +  4 

x(l  +  x^) 


15.    I    ~"    '   T    dx. 

jg      rV2x  +  3  dx. 


3x  -  2 

103.  Integrals  of  Powers  of  Trigonometric  Functions. 

(a)   I   sin"  X  cos"  x  dx  where  at  least  one  of  the  exponents  is  an 

odd  positive  integer.     This  includes   I   sin"»  x  dx  and  J  cos"  x  dx 
where  the  exponents  are  odd. 
Illustration  1. 

J  sin^  X  cos^  X  dx  =  J  (1  —  cos^  x)  cos*  x  sin  x  dx 

=  J  cos*  xsinx  dx  —  J  cos*  xsinxdx 
_       cos'  a;       cos^x        _, 
-  3-  +  -5— +  ^- 

Illustration  2. 

I  cos'  X  dx  =    1(1—  sin^  x)  cos  x  dx 

=    I  cos  X  dx  —  j  sin*  x  cos  x  dx 

sin'x   ,    _ 
=  sin  X  —  — o r  C. 

It  is  seen  that  the  process  consists  in  combining  one  of  the  func- 
tions sin  X  or  cos  x  with  dx  to  form  the  differential  of  —  cos  x  or 
of  sin  X,  respectively,  and  of  expressing  the  remaining  factors 
of  the  function  to  be  integrated  in  terms  of  cos  x  or  sin  z, 
respectively. 

13 


4.    I  sin*  X  dx 

6.    I  \/sin  X  cos'  x  dx. 


194  CALCULUS  [§103 

Exercises 

1.  fsin'zdx.  f  cos»a; 

•7.  7.       -y==  dx. 

2.  j  sin"  X  COS*  z  dr.  J  ^^^'^  ^ 

3.  I  cos*  X  sin'  x  dx.  g      I    ^'"'  ^     ^^^ 
r  .   .      .  t/  (cos^)' 

9.     I  sin*  a  cos'  a  da. 
6.     fcos*  X  sin'  x  dx.  10-  J  cos^  (2x+3)  sin'  (2x+3)  dx. 

(b)    I  sin"  z  COS"  x  dx  when  w  and  n  are  both  even  positive 

integers-     In  this  case  make  use  of  the  relations: 
sin'^  X  =  5(1  —  cos  2.t). 
cos^  X  =  5(1  +  cos  2x). 
sin  X  cos  X  =  ^  sin  2x. 
Illustration  1. 

I  sin'  X  dx  =    H(l  ~  cos  2x)  dx  =  \\dx  —  \\  cos  2x  dx. 
_  X       sin  2x 

~2      r^  +  ^- 

Illustration  2. 

I  sin'  X  cos'  X  dx  =  5  j  sin'  2x  dx  =  |  1  (1  —  cos  4x)  dx 

_  X       sin  4x 
~  8  32~  "^ 

Illustration  3. 

J  cos*  X  dx  =  J  I  (1  +  cos  2x)'  dx 

=  ij(l  4-  2  cos  2x  +  cos'  2x)  dx 

=  ix  +  4  sin  2x  +  I  I  (1  +  cos  4x)  dx 
=  f  ^  +  4  sin  2x  +  -^2  sin  4x  +  C. 
Illustration  4. 

J  sin'  X  cos*  X  dx  =  J  (sin  x  cos  x)*  cos*  x  dx 

=  I J  sin'  2x  (1  +  cos  2x)  dx 

=  iV J  (1  —  COS  4x)  dx  +  I  I  sin'  2x  cos  2x  dx 
=  iVx  —  Vv  sin  4x  +  Vif  sin'  2x  +  C. 


§103]  INTEGRATION  195 

Exercises 

1.  js'm'^xdx.  4.    jsm*xdx. 

2.  J  cos*  2x  dx.  5.    fsin*  3x  dx. 

3.  /  sin*  X  cos''  x  dx.  6.    j  cos*  5a;  dx. 

(c)  I  tan"  a;  rfx  and   I  cot"  x  da;. 
Illustration  1. 

J  tan^x  dx  =  J  tan^x  (sec^x  —  1)  dx  =      „      —  ftan^  x  dx 

tan' X       f/      «  .  V   , 

=  — 5 I  (sec'^x  —  1)  dx 

tan^x  ,        ,    /^ 

=  — o tan  X  -r  X  +  C. 

Illustration  2. 
J  cot'  X  dx  =  J  (csc^  X  —  ly  cot  X  dx 

=    I  esc*  X  cot  X  dx  —  21  csc^  x  cot  x  dx  +  I  cot  x  dx 
=  —  J  CSC*  X  +  csc^  X  +  log  sin  x  -\-  C. 

(d)  I  sec  X  dx  and  J  csc"  x  dx,  n  an  even  integer. 
Illustration  1. 

J  sec*  X  dx  =  J  (1  +  tan^  x)  sec'^  x  dx  =  tan  x  +  |  tan'  x  +  C. 
When  n  is  odd  this  method  fails.     (See  §106.) 

(c)    /  tan"  z  sec"  x  dx  and  J  cot*"  x  csc"  x  dx  when  n  is  a  positive 
even  integer,  or  when  m  and  n  are  both  odd. 
Illustration  1. 

J  tan*  X  sec*  x  dx  =  J  tan*  x(l  +  tan'^  x)  sec^  x  dx 
=  I  tan^x  +  7  tan^x  +  C. 
Illustration  2. 

J  tan'  X  sec'  x  dx  =  j  tan'^  x  sec^  x  sec  x  tan  x  dx 

=    I  (sec'^  X  —  1)  sec^  x  sec  x  tan  x  dx 
=   I  (sec*  X  —  sec'  x)  sec  x  tan  x  dx 
=  i  sec'  X  —  5  sec'  x  +  C. 


196 


CALCULUS 


I  §104 


If  m  is  even  and  n  is  odd  the  methods  of  §106  must  be  used,  for 
the  integral  reduces  in  this  case  to  the  integral  of  odd  powers 
of  the  secant. 

Exercises 

9.    I  tan^  X  sec'  x  dx. 
LO.    I  tan'  X  sec*  x  dx. 


1.  I  tan*  X  dx. 

2.  1  CSC*  X  dx. 

3.  1  tan^  X  sec'  x  dx. 

4.  I  tan*  X  sec'  x  dx. 
6.  I  cot*  X  dx. 

6.  I  csc«  X  dx. 

7.  I  tan*  X  sec'*  x  dx. 

8.  I  sec*  X  dx. 


I  tan  5  z  sec*  a;  dz. 
2.    I  (tan 2  x  +  tan*  x)  dx. 
1.    I  sec'  X  tan^  x  dz. 

4.  fctan  e  -f  cot  OY  dd. 

5.  jtan^ede. 
.    I  sec'  e  tan~*  6  d9. 

104.  Integration  of  Expressions  Containing  s/a.^  —  x%  Va*  +  x*> 
\/x2  —  a^  by  Trigonometric  Substitution.  The  methods  of  §103 
find  frequent  application  in  the  integration  of  expressions  which 
result  from  the  substitution  of  a  trigonometric  function  for  x 
in  integrals  containing  radicals  reducible  to  one  of  the  forms 
Va*  +  x\  Va*  -  x\  or  Vx*  -  a^. 

Illustration  1.  I  \/a^  —  x^  dx.  Let  x  =  a  sin  6.  Then 
dx  =  a  cos  0  d0,  and 

JVa^  -  a;2  dx  =  Ja^  cos^  0  dO  =  la^{e  +  |  sin  2d)  +  C 
=  ^a2(0  +  sip  0  cog  ^)  +  c 

=  la'  [sin-i  ^  +  ^,  aA^":^]  +  C 

=  W  sin-'  ^  +  JxVa*  -  x^  +  C. 
Illustration  2.     j  -\/a2  -|-  a;2  ^3  dx.    Let  x  =  a  tan  0.     Then 
J  Va»  +  x^x'dx  =a5  J  tan^  0  sec^  0  d9 

=  a'^Jtan^  6  sec^  0  tan  0  sec  0  d^ 
=  a»J(sec*  d  -  sec*  0)  tan  0  sec  &  dd 


§104] 


INTEGRATION 


197 


=  a6(i  sec*  d  -  I  sec3  6)  +  C 


a' 


1  + 


1  + 


Illustration  3. 


/ 


5  3 
5                         3 


+  (7. 


dx.     Let    X    =    a  sec  6.    Then 


dx  =  a  sec  d  tan  0  d0,  and 


sec5tan2  0d0 


sec  0 

=  a  Jtan2  6  dd 

=  a  (tan  9  -  0)  +  C 

=  a-*  /-;  —  1  —  a  sec"^  -  +  (7 

=  Vx^  —  a2  —  a  sec-i  -  +  C. 
o 

The  integration  can  also  be  performed  directly  if  the  numerator 
is  rationalized.     Thus, 

rVx2  -  a^  , 


(x2  —  a^)dx 


■  r-7^£=  -  a^  f 


dx 


=  \/x2  —  a2  —  a  sec-i  -  +  C. 
a 

The  substitutions  used  in  these  illustrations  are  summarized  in 
the  following  table: 


Radical 

Substitution 

Radical  becomes 

X  =  a  sin  ^ 
X  =  a  tan  9 
X  =  a  sec  d 

a  cos  9 
a  sec  9 
a  tan  9 

Va'  +x^ 
Vx^  -  a^ 

198  CALCULUS  [§104 


Expressions  involving  s/ax^  +  6x  +  c  can  frequently  be  inte- 
grated by  completing  the  square  under  the  radical  sign  and  making 
a  trigonometric  substitution. 

Illustration  1. 


r     xdx        r     xdx 

J  V3  +  2x-  x2      J  V'4  -  (x  -  1 


Let  X  -  1  =  2  sin  9.    Then  a;  =  1  +  2  sin  ^  and  dx  =  2  cos  6  dd. 
Hence 


/x  dx  r 

VS  +  2x  -  a;2  ""  ^ J 


(1  +  2  sin  6)  cos  Ode 


2  cos  0 
=  J(l  +  2  sin  0)d0 

=  0  -  2  cos  ^  +  C 

X  —  1 
=  sin-»  —2 VS  +  2x  -  x^  +  C. 

Illustration  2. 


0^]' 


/rfx r rfx__ 
\/(2aa;  -  x^y  ~  J  [a^  -  (x  -  a) 

Let  X  —  a  =  a  sin  ^.     Then  x  =  a(l  +  sin  d)  and  dx  =  a  cos  0  rf5. 

dx r  a  cos  5 

•\/"(2ax  -  x^y  ~~  J  a'coss  0 

=  ^Jsec^ede 


=  -„  tan  0  +  C 

^  1  sin  g 
a2  cos  0  "^  ^ 
X  —  a 
1         ~^ 


a2  1    , _ 

- V2aa;  -  x^ 


+  C 


o''  \/2ox  -  x2 


§105]  INTEGRATION  199 

Exercises 

dx 


dx 


1   r    dx           ^   r     dx  r 

2./>^'...    »./;vfc,  »-/a-.,vr^ 

3./xvrTT..x.   6.J^^7^,-  »./^^;r~r» 

10.  J  (a^  -  x^y-dx  =  3a^f  cos*  0  sm^O  de. 


Hint.     Let 


S  3       . 

xs  —  a^  sin''  e, 


X  =  a  sin^  0. 


4)5 
dx 


(16  -  x^)* 
dx 


r  x^dx  r    dx  r    dx 

'  J  Va^~^^'  '  J  xWx^^^'  ^-  J  a;(x»  - 

12.    fV9  -  5x2  dx.       14.    f{9  ~x^)^dx.         16.     1 

J  (x2  +  6x  +  25)  2  J  (x2  +  4x  -  5)^ 

Vl^^^"  20.  Jv'2  +  6x-x^cix. 

105.  Change  of  Limits  of  Integration.  In  working  the  pre- 
ceding exercises  by  substitution  it  was  necessary  to  express  the 
result  of  integration  in  terms  of  the  original  variable.  In  the 
case  of  definite  integrals  this  last  transformation  can  be  avoided 
by  changing  the  limits  of  integration. 


L.    I    x^Vo^ 


Illustration  1.    |    x^  Vo^  —  x^dx.    Let  x  =  a  sin  6.     Then 
dx=  a  cos  d  dd. 


When  X  =  0,  sin  0  =  0  and  ^  =  0. 

TT 

When  X  =  a,  sin  ^  =  1  and  ^  =  o* 


200  CALCULUS  [§105 

As  X  varies  continuously  from  0  to  o,  0  varies  continuously  from 

TT 

0  to  ^  •    Hence  we  have 


I    x^y/a^  -  x^dz  =  a^\  "sin^  Q  cos^  Q 
Jo  Jo 


dd 


=  a*{ld  -  A-  sin  Ad) 

2 

n 

wa* 
~  16 

r**     x'  dx 
Illustration  2.      1    — -. Let  x  =  a  tan  6.     Then 

Jo  Vfl^  +  x^ 

dx  =  a  sec^  6  dd. 

When  X  =  0,  tan  0  =  0  and  6  =  0. 

■r,^,                                            .               />               -                 1     /.              'T 

When  X  =  a,  tan  0  =  1  and  ^  =  7* 
As  X  varies  continuously  from  0  to  a,  0  varies  continuously  from 

IT 

0  to  T-    Hence  we  have 
4 

J^*     x'dx  C*  \~* 

-^===  =  a'  1    tan»  d  sec  d  dd  =  a^i  sec^  d  -  sec  d)\ 
0  VS^T^^  Jo  lo 

=  ia\2  -  V2). 

\/a2  —  x^  dx.     By    using    the  substitution 
X  =  a  sin  0  we  obtain 

/*2 

d5 


JVa^  -  x2  dx  =  a^  j    cos^  0 
0  «/0 


=  iaM^  +  ^  sin  20)r 


Tra'' 


l2 

~     4  * 

The  above  integral  is  of  frequent  occurrence  in  the  application 
of  the  calculus.     The  integrand,  \/a^  —  x^,  is  represented  graphic- 


§106]  INTEGRATION  201 

ally  by  the  ordinates  of  a  circle  of  radius  a,  center  at  the  origin. 
The  integral  then  represents  the  area  of  one-quarter  of  this  circle. 
(See  §§64  and  65.)  The  value  of  any  integral  of  this  form  may 
be  written  down  at  once.     Thus, 


I    V4  -{x-  5)2  dx  =    I    \/4  -  M^  rfw  = 
Jo  Jo 


7r22 


s: 


^^TT^    _ _  7r(3-hz^) 

\/3  +  z^  -  a;2  dx   =    -. 


Exercises 


1.     I    (9  -  a;*)*  dx.  7.     I 

Jo  Jo    (o^  +  X 

,.  f— i^.  s.  r 

Jo   V2ax-x'  J^A 

r  x^dx  g  r_ 

•  X  V9^r^  '  Jo  (^ 

,.  r     dx     .  ^,  r 


Vi"   X  (x^  -  4)' 
dx 


+  16)2 
dx 


x'^ix^  -  9)^ 


6 


I    ^25  -  x^dx.  11.    I    \/9  -  (x  -  4)2  dx. 

V^9  _  a;2  dx.  12.     I  Vb''  -  2/2  _  x2  dx. 

0  Jo 

106.  Integration  by  Parts.  The  differential  of  the  product  of 
two  functions  u  and  v  is 

d{uv)  =  udv  ■}-  V  du.  (1) 

Integrating  we  obtain 

uv  =    \udv  -\-  \vdu 
From  which 

r  u  dv  =  uv  —  J  V  du.  (2) 

This  equation  is  known  as  the  formula  for  integration  by  parts. 
It  makes  the  integration  of  u  dv  depend  upon  the  integration  of 
d»  and  of  v  du. 


202  CALCULUS  [§107 

Illustration  1.     j   x  log  x  dx.     Let    log  x  =  u   and   x  dx  =  dv. 
The  application  of  (2)  gives 

I  xlogx  dx  =  Ix^  log  X  —  I   I  x^-  dx 

=  |x2  log  X  -  lx^  +  C. 

Illustration  2.      I    xe'*da;.     Let  e^'dx  =  dv  and    x  =  u.     The 
application  of  (2)  gives 

I  xe^'dx  =  \xe^'  —  \   \  e^'  dx 
=  |xe3'  -  W  +  C 
=  ie3'(3x  -  1)  +  C. 

If  we  had  let  xdx  =  dv  and  e'*  =  w  we  should  have  obtained  a 
more  complicated  expression  to  integrate  than  that  with  which 
we  started. 

Exercises 

1.  I  x*  log  X  dx.  2.    j  X  cos  x  dx.  3.    I  sin~i  x  dx 

4.  I  x''  e*'  dx.     (Apply  formula  (2)  twice  in  succession.) 

6.  I  tan~i  X  dx.  9.    I  x  sin'  x  dx  . 

6.  I  X  sin  X  dx.  10.    I  log  x  dx. 

7.  I  x"  log  X  dx.  11.    I  x^  sin  2x  dx. 

8.  I  x'^  tan~^  2x  dx.  12.    I  sin  x  log  cos  x  dx. 

107.  The  Integrals  J    e"  sin  nx  dx,  J    e**  cos  nx  dx.    Let 

M  =  sin  nx  and  dv  =  €"dx.     Then 

Jc*  sin  nx  dx  =  -e»*  sin  nx (  e"'  cos  nx  dx. 
a                       a  J 

A  second  integration  by  parts  with  u  =  cos  nx  and  dy  =  e"dx 
gives 

f        .         ,        1        .  n  w*  f         .  . 

I  e"sm nxdx  =  -e<»*sin  nx ;e"cos  nx x  I  e"  sm  nx  dx. 

J  a  a^  a^  J 

The  last  term  is  equal  to  the  integral  in  the  first  member  multiplied 


§107]  INTEGRATION  203 


by  -^-    On  transposing  this  term  to  the  first  member  we  obtain 

^ —  I  e"^  sm  nx  ax  =  ^  (a  sm  nx  —  n  cos  nx)  +  C. 

Then 

/©ax 
e     sin  nx  dx  =  —^-r — i  (a  sin  nx  —  n  cos  nx)  +  C 

eax 

sin  (nx  -  a)  +  C,  (1) 


where 


and 


cos  a  = 


sm  a 


Va^  +  n2 
q 


The  student  will  show  in  a  similar  way  that 

/gax 
e"cos  nx  dx  =  ~ir~, — o  (n  sin  nx  +  a  cos  nx)  +  C 
a2  _i-  n2  V 


Va^  +  n2 
where 


a^  +  n^ 
eaz 

cos  (nx  -  a)  +  C,  (2) 


cos  a  = 


and 

n 


sm  a  = 


Va^ 


Exercises 

The  student  will  work  exercises  1-5  by  the  method  used  in  obtaining 
(1)  and  (2)  above.  In  the  remaining  exercises  he  may  obtain  the 
results  by  substituting  in  (1)  and  (2)  as  formulas. 

1.  fe-"  sin  7t  dt.  6.  \e-'^  cos  bt  dt. 

2.  fe-^  cos  St  dt.  7.  je-"-"  sin  ut  dt. 

3.  fe-o"  sin  St  dt.  8.  fc-o-^  cos  co<  dt. 

4.  fe-o-^'  cos  4<  d<.  9.  fe-"'"  cos  5t  dt. 

5.  je~''smxdx.  10.  j  e'"-' s'm  4:t  dt. 
11.  Find  a  in  exorcises  1-10. 


204  CALCULUS  [§109 

108.  J  sec^  X  dx.  This  integral  can  be  evaluated  by  a  method 
similar  to  that  used  in  the  last  article. 

/sec...x=/sec.sec..<*. 

=  sec  X  tan  x  —   \  sec  x  tan^  x  dx. 

Since  tan^  x  =  sec'^  a;  —  1, 

I  sec'  xdx  =  sec  x  tan  ^  —  j  sec'  x  dx  +  j  sec  x  dx. 

Transposing  the  next  to  the  last  term  to  the  first  member,  dividing 
by  2,  and  integrating  the  last  term  we  have 

I  sec'  X  dx  =  i  [sec  x  tan  x  +  log  (sec  x  +  tan  x)]  +  C. 

Exercises 

1.  I  csc^  X  dx.  5.    j  VoM-^  rfx. 

2.  jsec"  X  dx.  6.  J  Vx^  —  4x  +  11  (ia;. 

,.  dx.  7.    I         ,  dx. 

Va^  +  x2  dx.  8.    I    \/x2  —  9  dx. 

0  t/3 

109.  Wallis'  Formulas.  Formulas  will  now  be  derived  which 
make  it  possible  to  write  down  at  once  the  values  of  the  definite 
integrals: 

I    sin"  d  de, 


2 

COS"  6  dd, 
and 


f 


I    sm«  9  cos"  6  dd, 


where  m  and  n  are  positive  integers  greater  than  1. 

J    sin"  Odd  =   \     sin"-i  0 sin  d dd. 
0  Jo 


§109]  INTEGRATION 

Integration  by  parts  gives 


205 


sin"  ddd  =  —  sin»-^  d  cos  B 


a 


+  (n  -  1 )  I    sin"-2  e  cos2  B  dB 
2  0(1  -  sin2  B)  dB 


=  (n  —  1)  I    sin""' 

=  (n  -  1)  I     sin»-2  BdB  -  (n  -  1)  \     sin»  0 
Jo  Jo 

g  the  last  term  and  dividing  by  n 

J    sin"  BdB=  ^  ~     I     sin"-^  B 
«    Jo 


rfe. 


On  transposing  the  last  term  and  dividing  by  n  we  obtain 


dB. 


This  equation  can  be   regarded  as  a  reduction    formula  for 
expressing 


r 


sin"  B  dB 


in  terms  of  an  integral  in  which  sin  B  occurs  with  its  exponent 
diminished  by  2.     Applying  this  formula  successively  we  obtain 


I    sin"  BdB  = 1  I  ' sin"-"  B  dB 

Jo  n      n-  2X 


n  —  In  —  3n  —  5 
n      71  —  2  n  —  4 

(n-l)(n-3) 


■^BdB 


n{n  -  2) 
(n-l)(n-3) 


-^ (    sin  BdB'iin  is  odd. 


t 


n{n  -  2)    • 
[(n-l)(n-3) 


4-2 


^r 


dB 


sin"  0  dO 


4-2 


n(n  -  2)    •    •  •   31 
(n-l)(n-3)    •    •    •  S-Itt 
n(n  -  2)    •    •    •  4  2        2 


if  n  is  even. 

if  n  is  odd. 
if  n  is  even. 


206  CALCULUS  [§109 

From  the  fact  that  the  integrals 


f- 


sin"  X  dx 
Jo 
and 

''2 


£ 


COS" X  dx 


represent  the  areas  under  the  curves  y  =  sin"  x  and  y  =  cos"  x , 

IT 

respectively,   between  the  limits   x  =  0  and   x  =  ^,  it  is  clear 
from  the  graphs  that 


I     COS"  xdx  =  I 
Jo  Jo 


COS" xdx  =  \     sin"  x  dx. 


The  results  obtained  can  be  expressed  in  the  single  formula 

f  COS.  ...  ./W...  -  ^^^i^^^^^,-^^..) 

where  a  =  1  if  n  is  odd,  and  a  =  ^  if  n  is  even. 
In  a  similar  way  we  shall  evaluate 


T 

f: 


sin"  d  COS"  6  dd. 


I 


2  ri 

sin"  d  COS"  d  dd  =    |     sin^-i  ^  cos"  d  sin  ^  f/0 


sin*"-^  d  cos"+^  5 


m  —  _    . 
+  ^   ,    1    I      sm'"-^  d  cos"+2  0  d0 


n  +  1 

X 

^    I    ^    I      sm^-2  0  COS"  0(1  -  sm2  d)  dd 

1/^2  m  —  1    P 

J   I      sm^-^ecos"^^^  -  ^^-pj    I     siw-dcos^dde. 


n  + 


§109]  INTEGRATION  207 

Transposing  the  last  term  to  the  left  member  of  the  equation 


b-m]£ 


sin"  6  COS"  6  do  = 


sin"  6  COS"  6  dd 


m-1    C^    . 

I     ,      I        SI 

^  +  1  Jo 

m-1  r 


sin^-^acos"^^^ 


Apply  this  formula  successively  and  obtain 


r 


sin"  9  cos°  d  dd 


(w  —  1)  (w  — 


m  —  1       m—S 


m  -\-  n  m  -{-  n 


3_    p 

-2J0 


sm'"-2  0cos"ed^. 


sin"'"''^  COS"  0d^ 


{m  -\-  n)  {m  -\-  n 

(m-1)  (m-3) 


Z^pTV(^pcos«ed^    if 


m  IS  even 


(m  +  n)  (m  +  n  —  2) 
(m-i)(m-3)-  •  •i-(n-i)(n-3) 


■^ n 

•(^  +  3)J„ 


sin  6  COS"  ^  dd  if  mis  odd 


(in+n)(mH-n— 2) 
(m-i)(in-3)- 


•(n+2)(n)(n-2) 
i-(n-i)(n-3)- 


(m+n)(m+n— 2)  •  •  •  (n+2)(n)(n— 2) 
(m-i)(m— 3)  •  •  •  2 


IT.,        . 

if  n  is 


2  2 


if  n  is 
odd 


and 

m  is 
even. 


,     w      ,  N  /     ,    x/     ; — r    if  n  is  either  even  or 

(m+n)(m+n— 2)  •  •  •  (n+3)(n4-i)       j  •,       j      •      jj 

'     odd,  and  m  is  odd. 

The  right-hand  member  of  the  last  formula  of  this  group  can 
be  put  in  a  form  similar  to  the  others  by  multiplying  numerator 
and  denominator  by  (n  —  l)(n  —  3)  •    •    •  2  or   1.     It  becomes 

(m-i)(m-3)  •  •  •  2  •  (n-i)(n-3)  •  •  •  2  or  i 

(m+n)(m+n-2)  •  •  •  (n+3)(n+i)  (n-i)  (n-3)  •  •  •  2  or  i 

IT 

These  formulas  for     |      sin"*  ^  cos"  6  dd  can  all  be  expressed  in 
the  single  formula 

X 

r  sin-d  cos"d  dd  =  (m-i)(m-3)--2ori(n-i)(n-3)-2ori 

Jo  (m  +  n)  (m  +  n- 2)---2  or  I 


«• 


(2) 

IT 

where  a  =  1  unless  m  and  n  are  both  even,  in  which  case  a  =  „* 


208  CALCULUS  [§109 

Illustration  1.     By  formula  (1) 

_    8-6-4-2        128 


r 
r 

B; 


sin^  Odd 

9-7-5-31      315 

Illustration  2. 

X 

cos*edd  =  ^!!:  =  ^TT. 
4-2  2       16 

Illustration  3.     By  formula  (2), 

4.0.0         1 
sm^  X  cos'  X  dx  =  =  __. 

8-6-4-2      24 

Illustration  4. 


J  2     .  4-9'^-1  S 

sm^x  cos^a:  dx  =     ^     '^       =  _r_. 
0  9-7-5-31      315 


Illustration  5. 


sin'x  cos^a;  dx 


_   5-31-31  X       3t 


10-8-6-4-2  2      512 
Exercises 


cos"  X  dx. 


1.  I     sin^  e  de.  7.    I 
Jo  Jo 

2.  I     cos^^ede.  8.    I    sin^<^d<^. 

r 

3.  I     cos^ede.  9.    I     sin^  a;  COS*  X  dx. 


4.    I     sin'^ede.  10.    I     sin^  X  cos"  X  dx. 
Jo  Jo 

X  X 

6.    I     COS*  e  do.  11.    I     sin<  x  cos^  x  dx. 
Jo  Jo 

x^  !L 

6.    I    sin«fldfl.  12.    I    sii 
Jo  Jo 


sin'  <f>  cos  ff>  d4t. 


§110]  INTEGRATION  209 

13.  I     sin*  X  cos*  x  dx.  16.    I     x^  {a^  —  x*)    dx. 

14.  I     cos^  X  sin6  x  dx.  17.    I     (a*  -  x^)^^^. 
Jo  Jo 

15.  I     (a2-x2)^dx.  18.    I    x(a^  -  x^)^dx. 
Jo  Jo 

19.  j    a}{\  -  cos  e)HQ  =  4a2  j    sin^  |  dd. 

Let  0'  =  2-     Then  d0  =  2de'  and  «'  =  |  wnen  fl  =  tt,  and  fl'  =  0 
when  0  =  0.     Hence 

a}\     (1   -  coseyde  =  8a2  (     sin^O'dO'. 
Jo  •  Jo 

WaUis'  formula  can  now  be  applied. 

By  transformations  similar  to  the  foregoing  many  integrals  can  be 
put  into  a  form  to  which  Wallis'  formulas  can  be  applied. 

20.  I     cos*  2ede  =  \  I    cos*  e'  do'. 
Jo  Jo 

r  _ 

22.     I      x\/2ax  -  x*  dx. 


21.    I       (2ax  -  x*)^dx.     (Substitute  x  =  2a  sin*  6.) 

I 

»2a 


fl.  sixi  X    I    b  cos  X 

110.  Integration  of    f  — . — ^— dx.     Integrals  of  this 

'  c  sm  X  +  d  cos  X 


X 


form  can  be  reduced  by  the  substitution,  z  =  tan  g.     In  making 

this  substitution  it  is  necessary  to  express  sin  x,  cos  x,  and  dx  in 
terms  of  z.  This  is  easily  done  as  follows.  (The  student  is  ad- 
vised to  observe  the  method  carefully,  but  not  to  learn  the  results 
as  he  can  readily  obtain  them  whenever  needed.)     Since 

z  =  tan  ^> 

X  =  2  tan~*  z, 
14 


210  CALCULUS  [§110 

and 


dz 

dx  =  2~. — j — 7, 


Further, 

and 

Then 

and 


COS;^  = 


2  X  \  X       V'l+2=' 

sec  2       -y/l  +  tan2  o 


.      X  ^         X  X 

sin  7.  =  tan  7=.  cos  ^ 


2  2  ^"-^  2       VlT 


„   .    a;        a;  2z 

sin  X  =  2  sin  ^  cos  ^  = 


2        2       1+^2 


X  X       1  -  z^ 

cos  a;  =  cos^  7^  —  sin^  -  = 


'■/r 


2  2       1  +  z2 


dx 

Illustration  1.      I   ^ — r~j On   making  the    substitution 

'    ^  +  4  cos  X 


X 

z  =  tan  ^  we  obtain  by  using  the  values  just  found  for  cos  x  and 
dx  in  terms  of  z,^ 

2dz 


! 


l-\-z'       ^  2  '  ^^ 


1  +  4,    ,   _3 


^/r 


1  -z^  J  1  +2^  +  4(1  -z^) 

1   +2^ 

dz 


-/^ 


322 

2     C  \/3dz 


■J- 


VsJ  3z2  -  5 

2        ,„gV^i^-  +  c 


2\/3  \/5        \/3z  +  VS 

V3  tan  I  +  VS 

=  A  \/l5  log  ^ 3  +  C. 

\/3  tan  2  —  \/5 

1  The  student  will  derive  these  values  in  each  problem  worked  in  order  to  famil- 
iarize hinist'lf    with  the  method. 


§111]  INTEGRATION  211 

Illustration  2.    |  _ #:7— •     Let  z  =  tan  %     Then 

2dz 

dz 


f dx ^  r  i+g'  _  ^  r 

j5-3sinx       I  Qz      ~  ^J 


5  +  522  -  62 

1+02 

dz 


.  f     dz      _  ,  r 

V  .2  _  e,  +  1  -     J 


_     (2-|)^  +  ii 

=  f  •  I  tan-i  -^  +  C  =  ^  tan-i  ^^^  +  C' 

5  tan  2  ~  3 
=  §  tan-i -. +  C. 


Exercises 

The  student  will  find  cos  x,  sin  x,  and  dx  in  terms  of  the  new  vari- 
able in  each  of  the  exercises. 


.   r    dx  r 

^'  J  3  +  5  cos  x'  ^'  J  1 

i.    C l^- 7     fs 

J  5  —  3  cos  X  J  1 

,  r   ^-^   .  8 '  r  ^ 

J  4  —  5  sin  a;  *  J  sin  a 

/   sin  xdx  r 

•  J  2  +  sin  x'  ^-  J  : 

/COS  iC  1 

o  I  o 1^^-  10.   I 

3  +  2  cos  T  I  . 


—  3  sin  X 

+  4  sin  x 
+  2  sin  X 

+  sin  X 


dx 
dx. 

dx. 


x(l  +  cos  x) 
dx 


3  sin  2x 
dx 


4  —  5  cos  2x 


111.  Partial  Fractions.     A  rational  fraction  is  the  quotient  of 
two  polynomials,  e.g., 

ttox™  +  flix"*-^  +   •  •      +  am-ix  +  a„       4>{x) 


box"  +  bix"-^  +   •  •  •  +  b„-ix  +  6„        /(x) 


(1) 


1  The  integrand  is  not  in  the  form  given  in  the  heading  of  this  article,  but  the  sub- 
stitution z  =  tan  2  enables  us  to  transform  any  expression  containing  only  integral 

powers  of  sin  x  and  cos  x  into  a  rational  function  of  «,  i.e.,  into  a  function  containing 
only  integral  powers  of  z. 


212  CALCULUS  [§111 

If  the  degree  of  the  numerator,  m,  is  greater  than  or  equal  to  the 
degree  of  the  denominator,  n,  the  fraction  can  be  transformed  by 
division  into  the  sum  of  a  polynomial  and  a  fraction  whose  nu- 
merator is  of  lower  degree  than  the  denominator.  In  this  case 
the  division  is  always  to  be  performed  before  applying  the 
methods  of  this  section. 

The  integration  of  a  rational  fraction  cannot  in  general  be  aC' 
complished  by  the  methods  which  have  been  given  if  the  degree 
of  the  denominator  is  greater  than  2.  Illustrations  will  now  be 
given  of  a  process  by  which  a  rational  fraction  can  be  expressed 
as  the  sum  of  fractions  whose  denominators  are  either  of  the  first 
or  second  degrees. 

Illustration  1. 

x2  +  2 


/. 


dx. 


-2x2-9x4-18 
Factoring  the  denominator 

X'  -  2x2  -  9x  +  18  =  (x  -  2)(x  -  3)(x  +  3). 
Assume 

x2  +  2  A       ,_B_         C 


x3-2x2-9x  +  18      X -2^x -3"^x4-3' 

where  A,  B  and  C  are  to  be  so  determined  that  this  equation  shall 
be  satisfied  for  all  values  of  x.     Clearing  of  fractions 

x^  +  2  =  Ax2  -  9A  +  5x2  +  Bx  -  QB  +  Cx^  -  5Cx  +  6C 
=  {A+  B  +  C)x2  +  (B  -  5C)x  -  9A  -  65  +  6C. 

On  equating  the  coefficients^  of  x^,  x,  x",  we  obtain  the  following 
three  equations  for  the  determination  oi  A,  B  and  C. 

A-hB  +  C  =  1. 

5  -  5C  =  0. 

-  9A  -  6B  4-  6C  =  2. 

1  In  applying  this  process  use  is  made  of  the  fact  that  if  two  polynomials  in  x  are 
identically  equal,  the  coefBcients  of  like  powers  of  x  are  equal.  Thus,  given  the 
identity 

a«"  +  aia;"-l  +  •   •   •  +  ccn-i  x  +  a„  =  /Sox"  +  /3ii"-l  +  •   •   •  +0n-i  x  +  P„, 
then 

ao  =  Po 
ai  =  /Si 


§111]  INTEGRATION  213 

From  these  equations 

A=  -h 
5  =  V. 

C  =  U. 


Hence 


x^  +  2  -6  11  11 


a;3  _  2x2  _  9a;  _|_  18      5(3.  _  2)   '  6(x  -  3)  ^  30(a;  +  3) 
and 


r      x^+2         ^  r_rfx 

J  a;3  _  2x2  -  Qx  +  18  ^  ^J  X- 

=  -  l\og{x-2)  +  V  log  (x-3)+U  log  (x  +  3)  +  C. 


2 
+  V  Tr-.  +  U  r^ 


Short  Method.  The  foregoing  method  of  determining  the 
values  oi  A,  B,  •  •  •,  by  equating  coefficients  of  like  powers  of 
X,  is  perfectly  general.  However,  a  shorter  method  can  sometimes 
be  used.  Thus  in  the  illustration  just  given  write  the  result  of 
clearing  of  fractions  in  the  form 

a;2  +  2  =A(x-  S)ix  +  3)  +  5(x  -  2)(a;  +  3)  +  C(x  -2)(x-  3). 

Since  this  relation  is  true  for  all  values  of  x,  it  is  true  f or  x  =  2. 
On  setting  x  =  2,  we  obtain 

6  =  -  5A. 
Hence 

A  =  -i 
On  setting  x  =  3^  we  obtain 

11  =  6B. 
Hence 

B  =  V. 
On  setting  x  =  —  3,  we  obtain 

11  =  30(7. 
Hence 

/n*  _  -Li 

Illustration  2. 

x2  +  l 


/ 


(x  +  lKx-l)'^''- 


214  CALCULUS  [§111 

Let 

x'  +  l         _     A  B  C  D 

(x  +  l){x  -  1)3  ~  X  +  1  "•"  (x  -  1)3  +  (x  -  1)2  "^  a;  -  1* 

On  clearing  of  fractions, 

x^  +  I  =  Aix  -  ly  +  Bix  +  1)  +C{x-l){x-{-l)+D(x-iy{x+l), 

or 

x^  +  1  =  Ax^  -  3Ax2  +  3Ax-  A+  Bx  +  B  +  Cx^  -  C 

+  Dx^  -  Dx^  -  Dx  -j-  D. 
In  the  first  form  put  x  =  1.    Then 

5  =  1. 
In  the  first  form  put  x  =  —  1.    Then 
-  8A  =  2. 
Hence 

A=  -h 
Equating  coefficients  of  x'  in  the  second  form 

A  +  D  =  0. 
Hence 

D  =  -  A  =  I 

Equating  coefficients  of  x^  in  the  second  form, 

-SA  +  C  -  D  =  I. 
Hence 

O—    14-1-4    —    2' 

Consequently 

x^-\-l  ^      -1  1  1  1 

(x  +  l)(x  -  ly      4(x  +  1)  "*"  (x  -  ly  "^  2(x  -  1)2  ■+"  4(x  -  1) 
and 

J   (X  +  1)(X-1)3''''  ^Ja^  +  l+J   (.r_l)3 

=  _  ^  log  (X + 1 )  -  2^^^,  -  2(^  +  i  ^ 


§1111  INTEGRATION  215 

Illustration  3. 


/i 


3.r2  -  2a;  +  2 

ax. 


{x  -  l)(a:2-4x  +  13) 

Let 

3x2 -2a; +  2  ^  5.r  +  C 


(x -l)(a:2-4x  +  13)       x-1    '   x^  -  4a;  +  13 
Clearing  of  fractions, 

3x2  -  2x  +  2  =  A{x^  -  4x  +  13)  +  Bx(x  -  1)  +  C(x  -  1), 
or 

3x2  -  2x  +  2  =  Ax^  -  4Ax  +  13A  +  5x2  ^  Bx  +  Cx  -  C. 

In  the  first  form  put  x  =  1.     We  obtain 
3  =  lOA. 
Hence 

A  =  A. 

Equating  the  constant  terms  in  the  second  form, 

13A  -  C  =  2. 
Hence 

18  -  C  =  2 

and 

n  —  i-9 

O    —    1  u- 

Equating  the  coefficients  of  x^  in  the  second  form, 

A+  B  =  3. 
Hence 

5  =  3  —  "i  0  =  To . 
Consequently 

C        3x2-2x  +  2  .         3    r  rf^      ,    1    r27x  +  19 

J(x-l)(x2-4x  +  13)^^  =  '^"J^^  +  "^"Jx-2-4x  +  13''^ 

=  A  log  (x-1)  +  ,ij  ^._4^^i3  +  i3  J  (^^ 


dx 


-2)2  +  9 

X  —  2 
=  -h  log  (x-1)  +  2-&  log  (x2-4x+13)  +  H  tan-i  -3—  +  C 

Illustration  4. 

2x  dx 
(r+x)(l+x2)5 


Ji 


216  CALCULUS  [§111 

Let  2a; ^       Bx -\- C       Dx  +  E 

(1  +  x)(l  +  x^y  ~  1  +  a;  "•"  (1  +  x^y  "^  (1  +  x-")' 

In  Illustrations  1  to  4  a  fraction  was  broken  up  into  "partial 
fractions."  The  denominators  were  the  factors  of  the  denomina- 
tor of  the  given  fraction.  In  Illustrations  1  and  2  the  factors  were 
all  real  linear  factors,  while  in  Illustrations  3  and  4  there  were  also 
factors  of  the  second  degree  which  could  not  be  factored  into  two 
real  linear  factors.  The  method  of  procedure  will  be  further  indi- 
cated by  the  following  examples.  They  will  be  grouped  under  the 
numbers  I,  II,  III,  and  IV,  corresponding  to  Illustrations  1,  2,  3, 
and  4. 

I.  Factors  of  denominator  linear,  none  repeated. 

,  .  x'  +  5 ^  g  C 

^"^    {x  -  l)(x  +  l)(a;  -3)      a;-l"'"x  +  la;-3 

., .   x^  -I-  2a:  +  7 A_         B 

^'    (a;  +  4)  (2a;  4- 3)  (x- 2)  (3a; +  1)       a;  +  4  "*" 

+   ^ 


a;  -  2   '   3a;  +  1 
II.  Factors  of  denominator  linear,  some  repeated. 

a;^  +  2a;  +  5 ^_    .   ^_    .         C' 


(x-2)2(a;-3)'(x  +  l)       (a;  -  2)^   '   x-2    '    {x-ZY 

D  E 


(a; -3)2   '   a;  -3    '   x  +  1 


... x»  +  4a;  -  2 A  B 


(2x  -f  l)8(a;  +  3)(a;  -  4)^       (2a;  +  l)^  ^  (2a;  +  1)^ 


2x  +  1    '   x  +  S    '    (x  -  4)2   '   a;  -  4 

III.  Denominator   contains   factors   of   second   degree,    none 
repeated. 

x2  +  7x  +  3      _  Ax  +  B  C 

^^'    (x2  +  4)  (a;  -  2)  ~   x*  +  4   +  x  -  2* 

„ .   x»  -  3x  +  5 Ax-\-B         Cx-^D 

"'     /•-V.2  _i_  o^/•^2  _  /i^  j^  '7\/'^    1    o^  "~    ™2    I    o      I 


(x2  +  2)(x2  -  4x  +  7)(x  +  3)         x2  +  2    ^  x2  -  4x  +  7 

+  _^. 
X  -\-  3 

x2  +  2x  -  5      _  Ax  +  B C__  D 

^^^     (x2  +  7)(x-2)2        x2  +  7    "^  (x-2)2  +  x -2' 


§111]  INTEGRATION  217 

IV.  Denominator    contains   factors    of    second    degree,    some 
repeated. 

a;3  +  2x2  +  5  Ax  +  B 


(a) 


(x2  +  2x  +  10)2(x2  +  3)(a;  +  2)       (x^  +  2x  +  10)  ^ 

Cx  +  D  Ex-^F         G 


a;2  +  2x  +  10   '     x^  +  S     '  x  +  2 
Exercises 


X3  +  X  -  10  '^*'  *•    J    (i+1 

r  (x-4)^x  r 

J  X'  -  6x«  +  9x  °'  J 

rx«  +  x^  +  7x  +  1 


dx 

KxM-T)' 


(3  +  4x  -  x")  dx 
(x-  lKx''-2x  +  5)' 

5x2  _|_  13a;  _  7 


(x+4)(2x+l)2 


dx. 


dx 

I* 

+  x3+3 


—  9x 


dx.      (Divide  numerator  by  denominator.) 


CHAPTER  XIII 

APPLICATIONS     OF     THE     PROCESS      OF     INTEGRATION. 
IMPROPER  INTEGRALS 

112.  In  this  section  a  brief  summary  and  review  of  the  appHca- 
tions  of  the  process  of  integration  will  be  given. 

1.  Area  under  a  Plane  Curve:     Rectangular  Coordinates. 

nb 
A  =    I  f{x)dx. 


See  §64,  and  Fig.  46. 

2.  Area:    Polar  Coordinates. 


-'£ 


See  §99,  and  Fig.  74. 

3.  Length  of  Arc  of  a  Plane  Curve:     Rectangular  Coordinates. 

dx 


i:>F(i)'*- 


See  §69,  and  Fig.  49. 

4.  Length  of  Arc:    Polar  Coordinates. 


See  §98,  and  Fig.  72. 

5.  Volume  of  a  Solid  of  Revolution. 

»b 
F  =  I    Try"^  dx 


See  §68,  and  Fig.  49. 


218 


§112]  IMPROPER  INTEGRALS  219 

6.  Surface  of  a  Solid  of  Revolution. 


'x  =  b 

ds 


=  2ir  I    y 

Jx  =a 

y  ds. 


See  §70,  and  Fig.  49. 

7.  Water  Pressure  on  a  Vertical  Surface. 

P  =  k\    uz  du, 

Ja 

where  z  denotes  the  width  of  the  surface  at  depth  u  and  k  =  62.5 
pounds  per  cubic  foot  if  u  and  z  are  expressed  in  feet.  See  §72, 
and  Fig.  50. 

8.  Work  Done  by  a  Variable  Force.    See  §67. 

Exercises 

1.  Find  the  area  in^the  first  quadrant  between  the  circle  x^  +  y'^  =  a* 
and  the  coordinate  axes. 

The  definite  integral  which  occurs  in  the  solution  of  this  problem  is 
of  very  frequent  occurrence.     See  Illustration  3,  §105. 

2.  Find  the  area  bounded  by  the  lemniscate,  p^  =  a^  cos  20. 

3.  Find  the  length  of  one  quadrant  of  the  circle  x^  +  2/^  =  o*,  or 
X  =  a  cos  0,  y  =  a  sin  0. 

4.  Find  the  length  of  p  =  10  cos  0. 

6.  Find  the  volume  of  a  sphere  of  radius  a. 

6.  A  solid  is  generated  by  a  variable  square  moving  with  its  center 
on,  and  with  its  plane  perpendicular  to,  a  straight  line.  The  side  of 
this  square  varies  as  the  distance,  x,  of  its  center  from  a  fixed  point 
on  the  line,  and  is  equal  to  2  when  a;  =  3.  Find  the  volume  generated 
by  the  square  when  its  center  moves  from  a;  =  2  to  x  =  7. 

7.  Find  the  area  of  the  surface  of  a  sphere  of  radius  a. 

8.  The  unstretched  length  of  a  spring  is  25  inches.  Find  the 
work  done  in  stretching  it  from  a  length  of  27  inches  to  a  length  oi 


220  CALCULUS  (§112 

29  inches,  if  a  force  of  400  pounds  is  necessary  to  stretch  it  to  a 
length  of  26  inches. 

9.  A  trough  3  feet  deep  and  2  feet  wide  at  the  top  has  a  parabolic 
cross  section.  Find  the  pressure  on  one  end  when  the  trough  is 
filled  with  water. 

3  3  2 

10.  Find  the  length  of  the  curve  x*  +2/'  =a^,  orx  =  ocos'  6, 
y  =  a  sin'  B. 

11.  Show  that  the  work  done  by  the  pressure  of  a  gas  in  expanding 
from  a  volume  i^i  to  a  volume  V2  is  given  by 

I     p  dv. 

Jn 

where  p  is  the  pressure  per  unit  area. 

Hint.  Take  a  cylinder  closed  by  a  piston  of  area  A  forced  out 
a  distance  Ax  by  the  expanding  gas.  Denote  by  At/;  the  work  done 
by  the  gas  in  expanding  from  a  volume  r  to  a  volume  v  -\-  Av, 
Then, 

I] 

p  dv. 


-f 


12.  Find  the  area  of  one  quadrant  of  the  ellipse  x  =  o  cos  6, 
y  =  b  sin  6. 

13.  Find  the  area  of  one  loop  of  the  curve  p  =  a  cos  26. 

14.  Find  the  length  of  the  cardioid,  p  =  a(l  —  cos  6). 

15.  Find  the  volume  of  the  ellipsoid  of  revolution  generated  by 

revolving  the  ellipse  "i  +  j^i  ~  ^  about  the  X-axis;  about  the  F-axis. 

16.  A  volume  is  generated  by  a  variable  equilateral  triangle  moving 
with  its  plane  perpendicular  to  the  JC-axis.  Find  the  volume  of  the 
solid  between  the  planes  x  =  0  and  x  =  2,  if  a  side  of  the  triangle  is 
equal  to  2x^. 

17.  Find  the  area  of  the  surface  generated  by  revolving  about 
the  X-axis  the  portion  of  the  arc  of  the  catenary 


a 
^=2 


+  e 


■■] 


between  (0,  a)  and  (xi,  yi). 

18.  Find  the  area  under  one  arch  of  the  cycloid  x  =  a  {d  —  sin  d), 
y  =  a(l  —  cos  9). 


§112]  IMPROPER  INTEGRALS  221 

19.  Find  the  length  of  that  portion  of  9y^  =  x'  above  the  Z-axis 
between  x  =  0  and  x  =  3. 

20.  Find  the  volume  generated  by  revolving  the  portion  of  the 
catenary 


y 


« r  - .  --1 

2    e"   +e    « 


between  x  =  0  and  x  =  b  about  the  X-axis;  about  the  F-axis. 

21.  Find   the   volume   generated   by    revolving    the   hypocyoloid 

2  3  1 

x^   +  y^   =  o*,  or  X  =  a  cos^  0,  y  =  asm^  0,  about  the  X-axis. 

22.  Find  the  area  included  between  the  parabolas  47/^  =  25x  and 
5x^  =  \oy. 

23.  Find  the  area  between  the  X-axis,  the  curve  y  =  x''  —  4x  +  9, 
and  the  ordinates  x  =  1  and  x  =  7. 

24.  Find  the  area  between  the  curve  y  =  sin  x,  the  X-axis,  and 
X  =  0  and  x  =  w. 

26.  If  a  gas  is  expanding  in  accordance  with  Boyle's  law,  -pv  =  C, 
find  the  work  done  in  expanding  from  a  volume  vi  to  a  volume  V2. 
Represent  the  work  graphically  by  an  area. 

26.  Find  the  work  done  if  the  gas  is  expanding  in  accordance  with 
the  adiabatic  law,  py*  =  C. 

Hint.     From  the  result  of  Exercise  11, 


C 


k 


A-k   _  „.l-ft 


Vi^-'). 


Now, 

C   ==   PiVi*    =   ^2^2*. 

Hence 

W  =  j-iri;  (P2i'2  -  PlVl). 

Represent  the  work  graphically  by  an  area.     Use  the  same  scale  as 
in  Exercise  25. 

x^        V^ 

27.  Find  the  area  of  one  quadrant  of  the  ellipse  Ta  +  g"  ~  ■'■•     ^^ 

Exercise  1. 

28.  Find  the  length  of  p  =  e"*  from  ^  =  0  to  ^  =  27r. 

29.  Find  the  length  of  p   =  e""^  from  ^  =  0  to  (?  =  -  =" ,  if  a  is 
assumed  positive. 

30.  Find  the  area  bounded  by  the  cardioid  p  =  a(l  +  cosO). 

31.  Find  the  area  bounded  by  p  =  10  sin  d. 

32.  Find    the    area    bounded    by    the    hypocycloid  x  =  o  cos^  9, 
y  =  a  sin^  d. 


222  CALCULUS  [§112 

33.  Find  the  area  between  y^  =  Ax  and  y^  =  8x  —  x'^. 

34.  Find  the  work  done  by  a  gas  in  expanding  isothermally  from 
an  initial  volume  of  2  cubic  feet  and  pressure  of  7000  pounds  per  square 
foot  to  a  volume  of  4  cubic  feet. 

36.  Find  the  work  done  if  the  gas  expands  adiabatically.  Take 
k  =  \h  the  value  for  steam.     (See  Exercise  26.) 

36.  Find  the  pressure  on  a  trapezoidal  gate  closing  a  channel  con- 
taining water,  the  upper  and  lower  bases  of  the  wet  surface  being 
25  feet  and  18  feet,  respectively,  and  the  distance  between  them  being 
10  feet. 

37,  Find  the  area  between  the  catenary 


|[^ea+e    aj, 


the  X-axis,  and  the  ordinates  x  =  0  and  x  =  a. 

38.  Find  the  length  of  p  =  ad  from  0  =  0  to  fl  =  27r. 

39.  Set  up  the  integral  representing  the  length  of  one  quadrant 
of  the  ellipse  x  =  a  cos  0,  y  =  b  sin  0. 

40.  Find  the  volume  generated  by  a  circle  of  variable  radius  mov- 
ing with  its  plane  perpendicular  to  the  X-axis,  between  the  planes 
X  =  2  and  x  =  8.  The  radius  is  proportional  to  x^  and  is  equal  to  54 
when  X  =  3. 

41.  Find  the  volume  generated  by  revolving  one  arch  of  the  cycloid 
X  =  a{d  —  sin  d),  y  =  a(l  —  cos  6)  about  the  X-axis;  about  the 
tangent  at  the  vertex. 

42.  Find  the  area  of  the  surface  generated  by  revolving  a  quadrant 
of  a  circle  about  a  tangent  at  one  extremity. 

43.  If  the  density  of  a  right  circular  C3dinder  varies  as  the  distance 
from  one  base,  find  the  mass  of  the  cylinder. if  the  altitude  is  h  and  the 
radius  of  the  base  is  r. 

44.  The  force  required  to  stretch  a  bar  by  an  amount  s  is  given  by 

„       Eos 

where  E  is  the  modulus  of  elasticity  of  the  material  of  the  bar,  o  is  the 
area  of  the  cross  section,  and  L  is  the  original  length.  Find  the  work 
that  is  done  in  stretching  a  bar  whose  unstretched  length  is  400  inches 
to  a  length  of  401  inches,  if  £  =  30,000,000  pounds  per  square  inch  and 
o  =  1.5  square  inches. 

46.  Find  the  area  of  one  loop  of  p  =  10  sin  Zd. 


§113]  IMPROPER  INTEGRALS  223 

46.  Find  the  length  of 

y  =  ^\ 


|[^e-   +e~aj 


from  (0,  a)  to  {xi,  y,). 

47.  Find  the  length  of  one  arch  of  the  cycloid  x  =  a{d  -  sin.  e) , 
y  =  a{l  —  cos  e). 

48.  Find  the  volume  of  the  anchor  ring  generated  by  revolving  the 
circle  x^  +  (y  -  by  =  a^  about  the  X-axis,  a  being  less  than  b. 

a 

49.  Find  the  area  of  the  small  loop  of  p  =  a  sin^  ^  * 

o 

60.  Find  the  work  done  in  pumping  the  water  out  of  a  cistern  20  feet 
deep,  in  which  the  water  stands  8  feet  deep,  if  the  cistern  is  a  parabo- 
loid of  revolution  and  the  diameter  at  the  surface  of  the  earth  is  8  feet. 

51.  Find  the  volume  included  between  two  equal  right  circular 
cylinders,  radius  a,  whose  axes  intersect  at  right  angles. 

62.  Find  the  area  of  the  surface  generated  by  revolving  one  arch  of 
the  cycloid  x  =  a{e  —  sm  6),  y  =  a{l  —  cos  9),  about  the  X-axis; 
about  a  tangent  at  the  vertex. 

63.  Find  the  area  bounded  by  p  =3+2  cos  6. 

64.  Find  the  area  bounded  by  the  small  loop  of  p  =2  +  3  cos  B. 
66.  Find  the  area  of  the  surface  generated  by  revolving  the  cardioid 

p  =  a(l  +  cos  6)  about  the  polar  axis. 

56.  Find  the  volume  bounded  by  the  surface  of  Exercise  55. 

113.  Improper  Integrals.     Since  — ,  becomes  infinite  at 

■s/x  —  1 

X  =  1,  the  definite  integral 

r       1 

dx 


: 


■s/x 


must  not  be  evaluated  by  the  usual  process.     For,  the  assumption 
has  been  made  that  in  the  integral 


fix)  dx 


fix)  is  a  continuous  finite  function  at  x  =  a  and  x  =  6  as  well  as 
at  all  intermediate  points,  and  the  evaluation  of  this  integral  was 
based  on  the  area  under  the  curve  y  =  fix).    In  this  case 


224  CALCULUS  [§113 

becomes  infinite  at  the  lower  limit.     The  area  under  the  curve 

1 


y 


Vx^^ 


between  the  ordinates  x  =  1  and  x  =  7  has  no  meaning.  In  fact 
the  integral  in  question  has  no  meaning  in  accordance  with  the 
definition  of  a  definite  integral  already  given.  A  new  definition 
is  necessary.     We  define 

•^       1 

dx 


as 


r 


im  / 


dx, 

+,  Vx  -  1 

where  r/  is  a  positive  number,  if  this  limit  exists.     Otherwise  the 
integral  has  no  meaning.     Now, 

1™  f  -yL^  dx  -  >s,  (2Vi^i)  r 

Ji+,  y/x-\  I14-, 

=  I'i?  (2 V6  -  2v^)  =  2V6. 
Since  the  limit  exists  we  say  that 


J 


7       1 

■  dx  =  2\/6. 


Graphically  this  means  the  limit  as  17  approaches  zero  of  the  area 

under  the  curve  y  =      ,  between  the  ordinates  x  =  1  +  ?; 

V  a;  —  1 

and  X  =  7,  exists  and  is  equal  to  2\/6. 

Exercise  1,     Show  that 


i 


dx 


],  {X  -  1)' 
exists  if  0  <  w  <  \. 

On  the  other  hand,  when  n  =  1, 


im  r  1       ,         lim ,      /         ^sV 

Ji+v   X       I  ^  1^ 


1+17 


§113]  IMPROPER  INTEGRALS  225 

This  limit  does  not  exist  and  consequently  we  say  that 

1 


/ 


x-1 


dx 


has  no  meaning  or  does  not  exist. 

Graphically  this  means  that  the  area  under  the  curve 


y  = 


1 

X-  1 


between  the  ordinates  x  =  1  +  t]  and  x  =  7  increases  without 
limit  as  rj  approaches  zero. 
Exercise  2.     Show  that 


S. 


dx 


X    (^  -  1)" 

does  not  exist  if  n  ^  1.  (Note  that  the  case  n  =  1  has  just  been 
considered.)  If  n<0  no  question  as  to  the  meaning  of  the 
integral  can  arise.     Why? 

A  definite  integral  in  which  the  function  to  be  integrated 
becomes  infinite  at  the  upper  limit  is  treated  in  the  same  way. 
Thus 


I 


dx 


is  defined  as 

dx 


lim  p"" 


vn=- 


where  ?;  is  a  positive  number,  if  tliis  limit  exists. 
Exercise  3.     Show  that 


i 


dx 


0    (1-^)" 

has  a  meaning  in  accordance  with  this  definition  if  0  <  n  <  1,  and 
that  it  has  no  meaning  if  n  >  1.  If  »  <  0  no  question  can  arise  as 
to  the  meaning  of  the  integral. 

It  is  easy  to  see  how  to  proceed  in  case  the  function  under  the 

15 


226  CALCULUS  [§114 

integral  sign  becomes  infinite  at  a  point  within  the  interval  of 
integration.     Thus 

dx 

;;»  where  n  is  a  positive  integer, 


I 


is  defined  as 

limrr-"   dx__    r     dx 

where  r;  is  a  positive  number,  if  this  limit  exists.     If  not,  the 
integral  has  no  meaning.     Tf  n  <  0  no  limit  process  is  necessary. 

Exercises 
Evaluate  the  following  integrals  if  they  have  a  meaning: 

dx 


3x-4 
dx 

2 

I  3 

dx 

2)^ 


*  Jo   ^''  ■  Jo    VoT^'  •  J_,  X 

Jo     "^  X    V^^=^^  Jo     (X- 

4     f^^       dx  g     p       dx 

12.  Find  the  area  between  the  curve  y'  =  ^ — — — >  its  asymptote  and 

the  X-axis. 

114.  Improper  Integrals :  Infinite  Limits.  In  §113,  the  interval 
of  integration  was  finite.  In  other  words  neither  of  the  limits  oi 
the  integral 


fmdx 

was  infinite. 

The  integral 

p     dx 
Jo     ^'  +  «^ 

will  be  defined  as 

lim  r     dx 

6=  CO     1        3-2      1      ^ 

§114]  IMPROPER  INTEGRALS  227 

if  this  limit  exists.     Now 


lim 

b  = 


lim  1  ,     _,b       1  TT 

0       0-"  a  a       a  2 


im  r     d^       ^  ^^"^-tan-i- 
6™  I      2   I — 2  represents  graphically  the  limit  of  the  area  under 

the  curve  y  =  — ;rn — ;  between  the  ordinatcs  x  =  0  and  x  =  h 
"       x^  +  a^ 

as  &  increases  indefinitely. 

Consider 

dx 

X ' 

lim  /    dx       lim  lim  ,      , 


inci 

I 


X 


But  log  6  increases  without  limit  as  h  increases  witnout  limit. 
Hence   |     —  has  no  meaning. 


Exercises 
Evaluate  the  following  integrals  if  they  have  a  meaning: 

2.    I      e-'dx.  4.    I      x'^e-'dx. 

Jo  Jo 

6.  Find  the  area  between  the  witch,  ij  =    VXT^'  ^^^  ^^®  ^^^^  °^  ^* 


CHAPTER  XIV 


SOLID  GEOMETRY 


115.  Coordinate  Axes.  Coordinate  Planes.  Just  as  the  posi- 
tion of  a  point  in  a  plane  is  given  by  two  coordinates,  for  example 
by  its  perpendicular  distances  from  two  mutually  perpendicular 
coordinate  axes,  the  position  of  a  point  in  space  is  given  by  three 
coordinates,  for  example  by  its  perpendicular  distances  from  three 
mutually  perpendicular  planes  of  reference,  called  the  coordinate 
planes.  Let  the  three  coordinate  planes  be  those  represented  in 
Fig.  75,  viz.,  XOY,  called  the  ZF-plane,  YOZ,  called  the  FZ-plane, 
and  ZOX,  called  the  ZX-plane.  Jhen  the  position  of  the  point  P 
whose  perpendicular  distances  from  the  YZ-,  ZX-,  and  XF-planes 


/ 


Fig.  75. 


Fig.  76. 


are  2,  3,  and  1,  respectively,  is  represented  by  the  coordinates  2,  3, 
and  1.  The  lines  of  intersection  of  the  planes  of  reference  are 
called  the  axes.  Thus  X'OX,  Y'OY,  and  Z'OZ,  Fig.  76,  are  called 
the  axes  of  x,  y,  and  z,  respectively.  The  coordinates  of  a  point 
P  measured  parallel  to  these  axes  are  known  as  its  x,  y,  and  z 
coordinates,  respectively.  Thus  for  the  particular  point  P 
of  Fig.  lb,x  =  2,y  =  3,  and  z  =  1.  More  briefly  we  say  that  the 
point  P  is  the  point  (2,  3,  1).  In  general,  (x,  y,  z)  is  a  point  whose 
coordinates  are  x,  y,  and  z.     If  these  coordinates  are  given  the 

228 


§115]  SOLID  GEOMETRY  229 

position  of  the  point  is  determined,  and  if  a  point  is  given  these 
coordinates  are  determined. 

The  relation  between  a  function  of  a  single  independent  variable 
and  its  argument  can  be  represented  in  a  plane  by  a  curve,  the 
ordinates  of  which  represent  the  values  of  the  function  correspond- 
ing to  the  respective  values  of  the  abscissas.  Thus,  y  =  j{x)  is 
represented  by  a  curve.  To  an  abscissa  representing  a  given  value 
of  the  argument  there  correspond  one  or  more  points  on  the 
curve  whose  ordinates  represent  the  values  of  the  function.  In 
like  manner  a  function  of  two  independent  variables  x  and  y  can 
be  represented  in  space.  Choose  the  system  of  coordinate  planes 
of  Fig.  75.  Assign  values  to  each  of  the  independent  variables 
X  and  y.  These  values  fix  a  point  in  the  XF-plane.  At  this 
point  erect  a  perpendicular  to  the  XF-plane,  whose  length  z  repre- 
sents the  value  of  the  function  corresponding  to  the  given  values  of 
the  arguments.  Thus  a  point  P  is  determined.  And  for  all  values 
of  X  and  y  in  a  given  region  of  the  XF-plane  there  will,  in  general, 
correspond  points  in  space.  The  locus  of  these  points  is  a  surface. 
The  surface  represents  the  relation  between  the  function  and  its 
two  independent  arguments  just  as  a  curve  represents  the  relation 
between  a  function  and  its  single  argument. 

Thus  if  2  =  ±  V25  -x^  -  y^  =  /(x,y),  ±  \/l2  are  the 
values  of  the  function  corresponding  to  the  values  x  =  2  and 
2/  =  3.    Then  the  points  (2,  3,  2\/3)  and  (2,  3,  -  2V3)  lie  on  the 

surface  2  =  ±  V^25  -x^-  y\  If  x  = -3and?/ =  1,  2  =  ± -s/l5- 
The  corresponding  points  on  the  surface  are  (—  3,  1,  -v/lS) 
and  (-3,  1,  -  \/l5)- 

The  coordinate  planes  divide  space  into  eight  octants.  Those 
above  the  XF-plane  are  numbered  as  shown  in  Fig.  76.  The  oc- 
tant immediately  below  the  first  is  the  fifth,  that  below  the  second 
is  the  sixth,  and  so  on.  The  points  (2,  3,  2\/3)  and  (2,  3,  -  2\/3) 
lie  in  the  first  and  fifth  octants,  respectively.  The  points 
(-3,  1,  \/l5)  and  (  —  3,  1,  —  \/l5)  lie  in  the  second  and  sixth 
octants,  respectively. 

The  locus  of  points  satisfying  the  equation 

z  =    ±  V25  -  x^  -tj^  (1) 

iS  a  sphere  of  radius  5.     For,  this  equation  can  be  written  in  the 


230 


CALCULUS 


[§117 


form  x^  +  2/^  +  2^  =  25,  which  states  that  for  any  point  P  on  the 
surface  (1),  OP  =  Vx"^  +  y^  +  z^  =  5.  The  left  member  is  the 
square  of  the  distance,  OP,  of  the  point  P  (x,  y,  z),  from  0,  since 
OP  is  the  diagonal  of  a  rectangular  parallelopiped  whose  edges 
are  x,  y,  and  z.  If  then  the  coordinates  of  P  satisfy  (1),  this  point 
is  at  a  distance  5  from  the  origin.  It  lies  on  the  sphere,  of  radius  5, 
whose  center  is  at  the  origin. 

116.  The  Distance  between  Two  Points.     The   student  will 
show  that  the  distance  d  between  the  two  points  (xi,  yi,  Zi)  and 

(X2,  t/2,  Zz)  is  

d  =  V(x,  -  xi^  +  (i/2  -  yir  +  (22  -  zi)K  (1) 

See  Fig.  77.     If  the  point  (xi,  yi,  Zi)  is  the  origin,  (0,  0,  0),  the 


expression  for  d  becomes 
P  = 


VX2^    +  2/2^   +  22= 


(2) 


Fig.  77. 


Fig.  78. 


Exercises 

Find  the  distance  between  the  following  points : 

1.  (1,  2,  3)  and  (3,  5,  7). 

2.  (1,  -2,  5)  and  (3,  -2,  -1). 

3.  (0,  -3,  2)  and  (0,  0,  0). 

4.  (0,  0,  3)  and  (0,  2,  6). 
6.  (0,  0,-5  and  (2,  0,  6). 

6.  (-3,  2,  -1)  and  (0,0,  0). 

117.  Direction  Cosines  of  a  Line.  Let  OL,  Fig.  78,  be  any  line 
passing  through  the  origin.  Let  a,  jS,  and  y  be,  respectively,  the 
angles,  less  than  180°,  between  this  Hne  and  the  positive  direc- 
tions of  the  X-,   Y-,  and  ^-axes.     These  angles  are  called  the 


§118] 


SOLID  GEOMETRY 


231 


direction  angles  of  the  line,  and  their  cosines  are  called  the  direction 
cosines  of  the  line.  Let  P,  whose  coordinates  are  x,  y,  and  z,  be 
any  point  on  the  line.     Let  OP  =  p.     Then 

X  =  p  cos  a, 

y  =  p  cos  /3, 
and 

z  =  p  cos  y. 
Squaring  and  adding  the  above  equations  we  obtain 

2-2  _|.  y2  _^  2^  =  p2(cos^  a  +  cos^  /3  +  cos^  7). 
Since 

3.2  _j_   j^2   ^  2^    =    p2, 

cos^a  +  cos^/3  +  cos^Y  =  L  (1) 

The  direction  cosines  of  any  line  are  defined  as  the  direction 
cosines  of  a  parallel  line  passing  through  the  origin.  Then,  the 
sum  of  the  squares  of  the  direction  cosines  of  any  line  is  equal  to 
unity. 

Exercises 

Find  the  direction  cosines  of  the  lines  passing  through  each  of  the 
following  pairs  of  points. 

1.  (0,  0,  0)  and  (1,  1,  1). 

2.  (0,  0,  0)  and  (2,  -3,  4). 

3.  (0,  0,  0)  and  (-1,  2,  -3). 

4.  (1,  2,  3)  and  (5,  6,  7). 
6.  (-2,  3,  -1)  and  (-3,  -4,  3). 

118.  Angle  between  Two 
Lines.  Let  AB  and  CD,  Fig. 
79,  be  two  lines,  and  let  their 
direction  cosines  be  cos  aj,  cos  /3i, 
cos  7i,  and  cos  0:2,  cos  ^2,  cos  72, 
respectively.      Denote  the  angle 

between  the  lines  by  d.  Let  CH,  HK,  and  KD  be  the  edges  of 
the  parallelopiped  formed  by  passing  planes  through  C  and  D 
parallel  to  the  coordinate  planes.  The  projection  of  CD  on  AB 
is  clearly  equal  to  the  sum  of  the  projections  of  CH.  HK,  and 
KD  on  AB. 
Hence 

CD  cos  d  =  CH  cos  ai  +  HK  cos  /3i  +  KD  cos  71. 


232 


CALCULUS 


[§119 


Now 


and 


CH  =  CD  cos  a2, 

HK  =  CD  cos  /32, 


KD  =  CD  cos  72. 
Consequently 

CD  cos  0  =  CZ)(co8  ai  cos  ccz  +  cos  /3i  cos  /32  +  cos  71  cos  72). 
Hence 

cos  6  =  cos  ai  cos  0:2  +  cos  |8i  cos  182  +  cos  71  cos  72.      (1) 

Exercises 

Find  the  cosine  of  the  angle  between  the  lines  determined  by  the 
points  of  Exercises  1  and  2;  2  and  3;  3  and  4,  of  the  preceding 
section. 

119.  The  Normal  Form  of  the  Equation  of  a  Plane. — ^Let  ABC, 

Fig.  80,  be  a  plane.  Let  ON, 
the  normal  from  0,  meet  it  in 
N.  Let  the  length  of  ON  be  p 
and  let  its  direction  angles  be  a, 
j3,  and  7.  If  p,  a,  /3,  and  7  are 
given  the  plane  is  determined. 

We  seek  to  find  the  equation 
of  the  plane.  Let  P,  with  co- 
ordinates X,  y,  and  z,  be  any 
point  in  the  plane.  The  sum 
of  the  projections  of  OH  =  x,  HK  =  y,  KP  =  z,  and  PN  upon 
ON  is  ON  =  p. 

The  projection  of  OH  on  ON  is  x  cos  a. 
The  projection  of  HK  on  ON  is  y  cos  j8. 
The  projection  of  KP  on  ON  is  z  cos  7. 
The  projection  of  PN  on  ON  is  0. 
Hence 

.  X  cos  a  +  y  cos  ^  +  z  cos  7  =  p.  (1) 

If  P  does  not  lie  in  the  plane  ABC,  the  projection  of  PN  on  OiV 
is  not  zero,  and  the  coordinates  of  P  do  not  satisfy  (1).  Hence  the 
locus  of  a  point  satisfying  (1)  is  a  plane.  Equation  (1)  is  the 
normal  form  of  the  equation  of  the  plane,     p  is  taken  to  be 


§120]  SOLID  GEOMETRY  233 

positive.  The  algebraic  signs  of  cos  a,  cos/3,  and  cosy  are  de- 
termined by  the  octant  into  which  ON  extends. 

Illustration  1.     Find  the  equation  of  a  plane  for  which  p  =  2, 
a  =  60°,  13  =  45°. 

cos  a  =  ^, 

cos  /3  =  — -^. 
V2 
Then  by  (1),  §117, 

cos-  7  =  1  —  J  —  §• 
Hence 

cos  7  =  ±  i. 

The  equation  of  the  plane  is 

2  +  V2  -  2 

There  are  thus  two  planes  satisfying  the  conditions  of  the  problem, 
one  forming  with  the  coordinate  planes  a  tetrahedron  in  the  first 
octant,  the  other  a  tetrahedron  in  the  fifth  octant. 

Exercises 

1.  Find  the  equation  of  a  plane  if  a  =  60°,  /3  =  135°,  p  =  2,  and 
if  the  normal  ON  extends  into  the  eighth  octant. 

2.  If  a  =  120°,  /3  =  60°,  p  =  5  and  if  the  normal  ON  extends  into 
the  sixth  octant. 

120.  The  Equation  Ax  +  By  +  Cz  =  D.     The  general  equation 
of  the  first  degree  in  x,  y,  and  z  is 

Ax  +  By  ^  Cz  =  D,  (1) 

where  A,  B,  C,  and  D  are  real  constants.  D  may  be  considered 
positive.  For,  if  the  constant  term  in  the  second  member  of  an 
equation  of  the  form  (1)  is  not  positive  it  can  be  made  so  by 
dividing  through  by  —1. 


.(2) 


Divide  (1)  by  VA^  +  B^ 
A              ^  , 

+  C2 
B 

and  obtain 

D 

VA^  -\-  B^  +  C^ 

Va^ 

1 

+  B^ 
C 

+  C2^ 

^VI 

'  +  B- 

2  +  C2          \/A^ 

+  B^ 

+  C^' 

234  CALCULUS  [§121 

The  coefficient  of  x  is  either  equal  to  or  less  than  unity  in  numerical 
value.  It  can  then  be  considered  as  the  cosine  of  some  angle,  say 
a.  Similarly  the  coefficient  of  y  may  be  considered  as  the  cosine 
of  some  angle  /3,  and  that  of  z  as  the  cosine  of  some  angle  7. 
Further  the  sum  of  the  squares  of  these  coefficients  is  equal  to  L 
Hence  a,^,  and  7  are  the  direction  angles  of  some  line.  Then  (2) 
is  in  the  form 

X  cos  a  -^  y  cos  /3  +  2  cos  7  =  p,  (3) 

where 

and  cos  a,  cos  /3,  and  cos  7  are  the  coefficients  of  x,  y,  and  z,  respect- 
ively, in  equation  (2).  Hence  (3)  is  the  normal  form  of  the  equa- 
tion of  a  plane.  Equation  (1)  is  the  general  equation  of  the  first 
degree  in  the  variables  x,  y,  and  z.  Therefore  every  equation  of 
the  first  degree  in  x,  y,  and  z  represents  a  plane. 

Illustration  1.     Put    3x  —  2y  —  z  =  Q    in    the    normal    form. 
Divide  by  VZmTbM^  =  \/9  +  4  +  1  =  \/U  and  obtain 
3j 2y z__  _      6 

Vii     \/l4     Vii  ~  Vii 

The  plane  is     . —  units  distant  from  the  origin,  and  forms,  with  the 
Vl4 

coordinate  planes,  a  tetrahedron  in  the  eighth  octant. 

Exercises 

Transform  each  of  the  following  equations  to  the  normal  form,  find 
the  distance  of  each  plane  from  the  origin,  and  state  in  which  octant 
it  forms  a  tetrahedron  with  the  coordinate  planes. 

1.  Sx  -  2y  -  z  =  1.  6.  X  +  2?/  =  6. 

2.  X  +  y  +  z  =  —1.  7.  x  —  2  =  4. 

3.  X  -  3r/  +  2z  =  3.  8.  x  =  2. 

4.  x  -  2j/  +  3z  +  2  =  0.  9.  X  =  -1. 
6.  2x  -  y  -  z  -  1  =  0.  10.  z  =  y. 

121.  Intercept  Form  of  the  Equation  of  a  Plane.  We  seek  the 
equation  of  a  plane  whose  intercepts  on  the  X-,  Y-.  and  Z-axes 
are  a,  b,  and  c,  respectively. 


§121]  SOLID  GEOMETRY  235 

The  general  equation  of  a  plane  is 

Ax  +  By  -{-  Cz  =  D.  (1) 

The  constants  are  to  be  so  determined  that  the  plane  will  pass 
through  the  points  (a,  0,  0),  (0,  b,  0)  and  (0,  0,  c). 

On  substituting  the  coordinates  (a,  0,  0),  in  (1),  we  obtain 

Aa  =  D, 
or 

a 

Similarly,  since  (1)  passes  through  (0,  b,  0), 

Bb  =  D, 
or 

-?• 

And,  since  it  passes  through  (0,  0,  c), 

Cc  =  D, 
or 

C 

With  these  values  oi  A,  B,  and  C,  (1)  becomes 
a         0  c 


or 

a   '    b    '   c 


+  f  +  -  =  1-  (2) 


Equation  (2)  is  known  as  the  intercept  form  of  the  equation  of  a 
plane. 

Illustration.     Transform  the  equation  3x  —  2?/  —  52  =  4  to  the 
intercept  form.     Divide  by  4  and  obtain 

-+-^  +  -^=1 

3  ■^  5 

The  intercepts  on  the  X-,  F-,  and  Z-axes  are  i,  —2,  and  — |, 
respectively. 


236  CALCULUS  [§123 

Exercises 
Transform  each  of  the  following  equations  to  the  intercept  form  : 

1.  X  +y  +z  =  3.  4.  2a;  +  7y  -  3z  =  1 

2.  2x  -  3y  +  42!  =  7.  5.  x  -  y  +  Sz  =  -\. 

3.  2x  +  y  -  z  +  2  =  0.  6.  t/  -  2x  -  3z  =  5. 

122.  The  Angle  between  Two  Planes.  The  angle  between  two 
planes  is  the  angle  between  the  normals  drawn  to  them  from  the 
origin.  The  cosine  of  the  angle  between  the  normals  can  be  found 
by  formula  (1)  §118,  in  which  ai,  /3i,  71  and  0:2,  ^2, 72  are  the  direc- 
tion angles  of  the  normals. 

Illustration.     Find  the  angle  between  the  planes 

x  +  y  +  z  =  l  (1) 

and  2x  +  y  +  2z  =  S.  (2) 

Transform  these  equations  to  the  normal  form  and  obtain 

Vs     Vs    V^     Vs 

and 

2a;       V      22 

3+1+3=1.  (4) 

The  direction  cosines  of  the  normals  to  the  first  and  second 

planes  are  —t=,  —j=,  — ^,  and  f ,  \,  f ,  respectively.     Then,  if  Q  is 

the  angle  between  the  normals,  formula  (1),  §118,  gives 

fl  2,1,2  5 

cos  Q  =  - — 7=  +  - — 7^  + 


3\/3      3\/3      3\/3      3\/3 
From  which  Q  =  74.5°. 

•    Exercises 
Find  the  angle  between  the  following  pairs  of  planes : 

1.  X  -  3j/  +  2z  =  6  and  x  -  2?/  +  2  =  1. 

2.  X  -  2?/  +  3z  =  2  and  2x  +  y  -  2z  =  3. 

123.  Parallel  and  Perpendicular  Planes.     If  two  planes  are 
parallel  0  =  0  and  cos  Q  —  \.     If  they  are  perpendicular  Q  =  90° 
and  cos  6  =  0. 
Let 

Arx-^  B,y  +  C,z  =  Di  (5) 

and 

AiX  +  B^y  +  C2Z  =  D2  (6) 


§124]  SOLID  GEOMETRY  237 

be  the  equations  of  two  planes.     After  writing  these  equations  in 
the  normal  form  it  is  found  that 

AiA2  +  BiB2-\-CiC2 
cos  d  =  , (7) 

If  AxA^  +  B,B^  +  C1C2  =  0,  (8) 

cos  B  =  Q  and  the  planes  (5)  and  (6)  are  perpendicular. 

If  the  planes  (5)  and  (6)  are  parallel,  the  corresponding  coeffi- 
cients must  be  equal  or  proportional.  For  then  and  only  then  will 
their  normals  be  parallel. 

Exercises 

From  the  following  equations  pick  out  pairs  of  parallel  planes  and 
pairs  of  perpendicular  planes. 

1.  a;  +  J/  +  z  =  6. 

2.  X  —  y  -  z  =  2. 

3.  2x  +  2y  +  2z  =  7. 

4.  3x  -  27/  -  z  =  8. 
6.  2x  -  3i/  +  z  =  1. 

124.  The  Distance  of  a  Point  from  a  Plane.  Let  {xi,  iji,  Zi) 
be  any  point  and  let 

Ax  +  By  +  Cz  =  D 
be  the  equation  of  a  plane.     We  shall  find  the  distance  of  the 
point  from  the  plane. 
Now 

Ax  +  By  +  Cz  =  K, 
where  K  is  any  constant,  is  the  equation  of  a  plane  parallel  to  the 
given  plane.  (See  §123.)  Let  us  choose  K  so  that  this  plane  shall 
pass  through  the  given  point  (xi,  ?/i,  Zi).  To  do  this  substitute 
the  coordinates  of  the  point  in  the  equation  and  solve  for  K. 
This  gives 

K  =  Axi  +  Byi  +  Czi. 
Placing  the  equation  of  each  plane  in  the  normal  form  we  have 

Ax-hBy  +  Cz      D 


and 

R 

Ax  +  By  +  Cz       K 
R                  R  ~ 

R 

Axi  +  Byi  +  Czi 
R 

where  R 

=  VA^  +  B^  +  C. 

238 


CALCULUS 


[§125 


The  given  plane  is  -h  units  distant  from  the  origin,  and  the  plane 

through  the  point  (xi,  yi,  Zi)  is p — units  distant 

from  the  origin.     Then  the  distance,  d,  between  the  two  planes, 
and  hence  the  distance  of  the  given  point  from  the  given  plane, 
is  equal  to  the  difference  of  these  two  distances,  or 
d  =  Axi  +  Byi  +  Czi  -  D 

\/A24^B2"+C2 

Illustration.     Find  the  distance  of  the  point  (1,  2,  —1)  from  the 
plane  3x  —  y  +  z-{-7  =  0. 

Axi  +  Byi  -\-Czi-D  _  31  -  1-2  +  !•(  -  1)  +7^      7 

Vn 


d  = 


VA^  +  B^  +  C^ 


V32+(-l)2  +  P 


Exercises 


In  each  of  the  following  find  the  distance  of  the  given  point  from 
the  given  plane: 

1.  (3,1,-2);        3x  +  y  -  2z  -  6  =  0. 

2.  (-1,2,-3);    x-y  -2z  +  l  =0. 

3.  (0,  2,-3);       2x  +  3y  -  5z  -  10  =  0. 

125.  Symmetrical  Form  of  the  Equations  of  a  Line.     Let  PPi, 

Fig.  81,  be  a  line  passing  through 
the  given  point  Pi  (xi,  yi,  zi), 
and  having  the  direction  cosines 
cos  a,  cos  /3,  cos  y.  In  order  to 
find  the  equations  of  the  line, 
let  P  (x,  y,  z),  be  any  point  on 
the  line  and  denote  the  distance 
PPi  by  d.     Then 


Fig.  81. 
and  therefore 


Xi 


X  —  Xi 

=  d  cos  a, 

y  -  2/1 

=  d  cos  /3, 

Z   —  Zi 

=  d  cos  7, 

y  -  2/1 

Z  -  Zi 

cos  j3 

cos  7 

(1) 

COS  a  — 

These  equations  are  known  as  the  symmetric  equations  of  the 
straight  line. 


§125]  SOLID  GEOMETRY  239 

Frequently  a  straight   line  is  represented    by    the  equations 
of  two  planes  of  which  it  is  the  intersection. 
Illustration  1. 

3x-y  +  l=0,  (2) 

5x  -z  =  S.  (3) 

From  these  equations  the  symmetrical  form  of  the  equations  can 
readily  be  obtained.     From  (2)  and  (3)  we  obtain 
_y-l _z+3 

"^  ~      3      -  ^5~' 
or 

x-0       y-1       z  +  3  ... 

nr  =  ~3~  =  ~5-  (^> 

The  denominators,  1,  3,  and  5,  of  (4)  are  not  the  direction  cosines 
of  the  line,  but  they  are  proportional  to  them.  Upon  dividing 
each  by  \/35,  the  square  root  of  the  sum  of  their  squares,  they 
become  the  direction  cosines.     Then 

X  -0  _  y-1  _  2  +  3 
1       ~      3     ~     5 


-s/35         \/35      VS5 
is  the  symmetrical  form  of  the  equations  of  the  line. 

The  line  therefore  passes  through  the  point  (0,  1,-3)  and  has 
the  direction  cosines  given  by  the  denominators  in  the  preceding 
equations. 

Illustration  2.  Consider  the  line  which  is  the  intersection  of 
the  planes 

13a:  +  52/  -  42  =  40, 
-13x  +  lOy  -2z  =  23. 
On  eliminating  x  we  obtain 

5y  -  2z  =  21, 
and  on  eliminating  y  we  obtain 

13x  -2z  =  19. 
From  the  last  two  equations  we  find 

5y  -  21       13x  -  19 

'  =  —2~  = 2 ' 

or 

x-n      y-\^      z  -  0 


A  i  1 


240 


CALCULUS 


[§126 


These  are  the  equations  of  a  line  which  passes  through  the  point 
(+3,  ^6*",  0)  and  whose  direction  cosines  are  proportional  to  A, 
5,  and  1.     The  student  will  find  the  direction  cosines. 

In  Illustration  1,  equation  (2)  represents  a  plane  parallel  to  the 
^-axis  whose  trace  in  the  XF-plane  is  the  line  3x  —  7j  -{-  1  =  0. 
Equation  (3)  represents  a  plane  parallel  to  the  F-axis  whose  trace 
in  the  ZX-plane  is  the  line  5x  —  z  =  3. 

In  Illustration  2  the  position  of  the  two  planes  which  intersect 
in  the  straight  line  is  not  so  evident.  By  eliminating  first  x  and 
then  y,  the  equations  of  two  planes  passing  through  the  same  line 
are  obtained,  one  of  which  is  parallel  to  the  X-axis  and  the  other  to 
the  y-axis. 


Exercises 
Put  the  equations  of  the  following  lines  in  the  symmetrical  form : 

1.  X  +  2y  +3z  =6, 
X  —    y  —    2  =  1. 

2.  a;  +    y  —    z  =  1, 
a;  -  3y  +  2z  =6. 

Z.  X  -    y  +  2z  =  0, 

X  +  2y  -  3z  =0. 

126.  Surfaces  of  Revolution.    Let 


2/2  =  42 


(1) 


Fig.  82. 


be  the  equation  of  a  curve  in  the 
FZ-plane,  Fig.  82,  and  let  it  be 
rotated  about  the  Z-axis.  The  sur- 
face generated  is  a  surface  of  revolu- 
tion. Any  point  D  on  the  curve  de- 
scribes a  circle  of  radius  CD,  equal  to 
the  7/-co6rdinate  of  the  point  D. 
During  the  revolution  the  z-co6rdinate 
does  not  change.  Let  P  be  any  posi- 
tion taken  by  D  in  the  revolution.  Let 
the  coordinates  of  P  be  (x,  y,  z). 


But  by  (1), 


x2  -f  7/2  =  icpy  =  {coy 

{CDY  =  iz, 


(2) 


§126]  SOLID  GEOMETRY  241 

where  z  is  the  common  ^-coordinate  of  Z)  and  P.  Then  (2) 
becomes 

x2  +  2/2  =  42,  (3) 

an  equation  satisfied  by  any  point  on  the  surface  of  revolution. 
We  note  that  (3)  is  obtained  from  (1)  by  replacing  ^/^  by  x^  +  y"^, 
or  y  by  -s/x^  +  y"^. 
In  general,  if 

/(y,  2)  =  0  (4) 

is  the  equation  of  a  plane  curve  in  the  FZ-plane,  the  equation  of 
the  surface  of  revolution  generated  by  revolving  it  about  the  Z- 
axis  is  obtained  by  writing  -s/x^  +  y"^  for  y,  i.e.,  the  equation  of 
the  surface  of  revolution  is 

fiVx^TV',  ^)  =  0-  (5) 

This  equation  can  also  be  regarded  as  the  equation  of  the  surface 
generated  by  revolving  the  curve  f{x,  z)  =  0,  lying  in  the  XZ- 
plane,  about  the  Z-axis. 

Similarly,  /(i/,  V^M^)  =  o  (6) 

is  the  equation  of  the  surface  generated  by  revolving  the  plane 
curve /(y,  x)  =  0  about  the  F-axis;  and 

<A(x,  V^M=^)  =  0  (7) 

is  the  equation  of  the  surface  generated  by  revolving  the  plane 
curve  0(.T,  2)  =  0  about  the  X-axis. 

Illustration  1.     The  equation  of  the  surface  generated  by  rotat- 
ing x^  -{•  iy  —  ^y  =  a^  about  the  X-axis  is 

x^  +  [Vy^  +  2'  -  ^Y  =  a^ 

Exercises 

Find  the  equation  of  the  surface  generated  by  rotating: 

1.  y  =  x^  about  the  F-axis. 

2.  y  =^  x^  —  a^  about  the  X-axis. 

3.  6^x2  +  a^yi  =  a^b^  about  the  X-axis. 

4.  6^x2  -  0^2/2  =  a^b^  about  the  X-axis. 
6.  &2x2  —  a-y^  =  a-b^  about  the  F-axis. 

6.  x2  +  y^  =  a^  about  the  F-axis. 

7.  x2  -f-  t/2  =  a2  about  the  X-axis. 

8.  y  =  mx  about  the  X-axis, 

9.  J/  =  mx  about  the  F-axis. 

16 


242  ■         CALCULUS  [§127 

127.  Quadric  Surfaces.     Any  equation  of  the  second  degree 
between  x,  y,  and  z,  of  which 

Ax^  +  By^  +  Cz^  +  Dxy  +  Eyz  +  Fxz  -\- Gx  -{-  Hy  +  Kz  -h  L  =  0 

(1) 
is  the  general  form,  represents  a  surface  which  is  called  a  quadric 
surface,  or  conicoid. 

By  a  suitable  rotation  and  translation  of  the  axes,  the  equation 
of  any  quadric  surface  can  be  put  in  one  of  the  following  forms: 
•  x^      v^      z^ 

±  ^'  ±  P  ±  ^  =  1'  (2) 

rp2  y2  2*2 

<^  ±  P  ±  C-'  =  »■  (3) 

X^  7/2 

-,±i,=  +  2c2.  (4) 

The  particular  form  assumed  by  the  equation  depends  upon  the 
values  of  the  coefficients  in  (1). 
The  quadric  surface 

X^  0/2  2j2 

^=  +  f=  +  ,T  =  l  (5) 

is  called  the  ellipsoid.  To  find  the  shape  and  properties  of  this 
surface,  let 

X  =  k,  (6) 

where  k  is  any  real  constant.  This  equation  represents  a  plane 
perpendicular  to  the  axis  of  x.  Equations  (5)  and  (6)  considered  as 
simultaneous  equations  represent  the  curve  of  intersection  of  the 
ellipsoid  with  the  plane.  If  x  is  eliminated  between  (5)  and  (6) 
there  results 

t/2  ^2 


/by/a^-k^y  ^  /cVa^  -  k^V 


the  equation  of  the  curve  of  intersection  in  the  plane  x  =  k. 
Equation  (7)  is  the  equation  of  an  ellipse.     The  semi-axes  of  the 

„.  &\/a2  -  k^       ,  c\/a^  -  k^      rru  u    * 

ellipse  are  —^ and  —^ .     These  axes  grow  shorter 

a,  a 

as  A;  increases  in  numerical  value  from  0  to  a.     When  k  =  ±  a 


§127]  SOLID  GEOMETRY  243 

the  elliptical  section  reduces  to  a  point.  When  |  A;|  >  a,  the  lengths 
of  the  axes  of  the  ellipse  become  imaginary,  i.e.,  the  plane  x  =  k, 
(,\k\  >  a),  does  not  meet  the  surface  (5)  in  real  points.  Hence  the 
surface  is  included  between  the  planes  z  =  +  a. 

The  above  discussion  shows  that  the  surface  represented  by  the 
equation  (5)  is  included  between  the  planes  x  =  +  a;  is  symmet- 
rical with  respect  to  the  FZ-plane;  and  has  elliptical  sections 
made  by  planes  perpendicular  to  the  axis  of  x.  These  sections 
grow  smaller  as  the  cutting  plane  is  moved  away  from  the  YZ- 
plane  and  at  a  distance  +  a  reduce  to  a  point. 

In  a  similar  manner,  by  taking  y  =  k,  and  then  by  taking 
z  =  k,  the  student  will  discuss  plane  sections  of  the  ellipsoid  (5) 
perpendicular  to  the  F-axis  and  to  the  Z-axis. 

a,  b,  and  c,  are  called  the  semi-axes  of  the  ellipsoid. 

It  can  be  shown  that  any  plane  section  of  the  ellipsoid  is  an 
ellipse. 

The  surface  represented  by 


/2 


^^ .  r  _  1  _  1  rsi 

^2  +  ^,2    c^  -  1  -  w 

will  now  be  discussed.    Let  z  =  k.     Then 

S  +  f:  =  i  +  l'  (») 

is  the  equation  of  the  plane  section  made  hy  z  =  k.     It  is  an 

ir           .                ■                  aVcM^2        ,  bVc^  +  k^       rp, 
ellipse  whose  semi-axes  are  — ' and  — ^^ ■ .      They 

increase  in  length  with  the  numerical  value  of  k.  The  axes 
have  a  minimum  length  when  k  =  0.  The  surface  represented  by 
equation  (8)  is  symmetrical  with  respect  to  the  XF-plane,  and 
every  section  parallel  to  this  plane  is  an  ellipse.  The  smallest 
elliptical  section  is  that  made  by  the  XF-plane. 
li  X  =  k,  equation  (8)  becomes 

P  -  c"^  =  1  -  a^'  (10) 

an  hyperbola. 

If  A;  <  a,  the  transverse  axis  of  the  hyperbola  is  parallel  to  the 


244  CALCULUS  [§128 


F-axis.     If  fc  >  o,  the  transverse  axis  is  parallel  to  the  Z-axis. 
When  k  =  a,  equation  (10)  reduces  to 

or 


r    z'^^^ 


(^+^)(i-:-)=». 


the  equation  of  two  straight  lines. 

The  student  will  discuss  the  curves  of  intersection  of  the  surface 
(8)  with  planes  parallel  to  the  XZ-plane. 

The  surface  is  called  the  hyperboloid  of  one  sheet,  or  of  one 
nappe. 

Exercises 

The  student  will  discuss  the  following  surfaces  and  make  sketches 
of  them: 

a;2       y2        z2 

1.  —2~  hi 2  ~  ^»      ^^^  hyperboloid  of  two  sheets. 

2.  ~^—  Ti  =  2cz,  the  hyperbolic  paraboloid. 

-j;2  y2 

3.  — 5  +  r^  =  2c2,  the  elliptic  paraboloid. 

128.  Cylindrical  Surfaces.     If  the  circle 

x2  +  2/2*=25  (1) 

be  moved  parallel  to  itself  so  that  all  of  its  points  describe  lines 
parallel  to  the  Z-axis,  it  will  generate  a  right  circular  cylinder. 
The  equation  of  this  cylinder  is  sought.  In  any  plane  z  =  k,  the 
relation  between  x  and  y  for  points  in  the  curve  of  intersection  of 
this  plane  and  this  cylinder  is  the  same  as  that  for  points  in  the 
plane  z  =  0,  viz.,  x"^  -{-  y^  =  25. 

Now,  this  equation  is  satisfied  by  all  points  on  the  surface  for 
all  values  of  z.     Hence  it  is  the  equation  of  the  surface. 

The  cylindrical  surface  just  considered  can  be  regarded  as 
generated  by  a  line  moving  parallel  to  the  Z-axis  and  passing 
through  points  of  the  circle  x^  -\-  y^  =  25  in  the  plane  z  =  0. 

In  general  a  cylindrical  surface  is  a  surface  generated  by  a  line 
moving  parcdlel  to  itself. 


§128]  SOLID  GEOMETRY  .  245 

It  is  clear  that  the  equation 

Kx,  y)  =  0  (2) 

represents  the  cylindrical  surface  generated  by  a  line  moving 
parallel  to  the  Z-axis  and  passing  through  points  of  the  curve 
f{x,  y)  =  0  in  the  plane  z  =  0.     The  equation  of  a  section  of  (2) 
made  by  any  plane  z  =  A;  is  j{x,  y)  =  0. 
Thus 

$  +  $-^  (3) 

represents  an  elliptical  cylinder  whose  elements  are  parallel  to  the 
Z-axis. 

3.2  _j_  1^2  _  2ax  (4) 

represents  a  circular  cylinder  whose  elements  are  parallel  to  the 
Z-axis.  The  center  of  the  section  in  the  plane  z  =  0  is  the 
point  (a,  0). 

By  the  same  reasoning 

7/  +  z^  =  a^  (5) 

represents  a  circular  cylinder  whose  elements  are  parallel  to  the 
X-axis. 

z2  =  4x  (G) 

represents  a  parabolic  cylinder  whose  elements  are  parallel  to  the 
y-axis. 
The  plane 

a;  -  4y  +  3  =  0  (7) 

can  be  regarded  as  a  cylindrical  surface  whose  elements  are 
parallel  to  the  Z-axis  and  which  pass  through  the  line 

X   ,   3 

y-i  +  i 

in  the  plane  z  =  0. 

In  general,  an  equation  in  which  one  of  the  letters  x,  y,  z  is 
absent,  represents  a  cylindrical  surface  whose  elements  are 
parallel  to  the  axis  corresponding  to  the  letter  which  does  not 
appear  in  the  equation. 


246 


CALCULUS 


I§129 


Exercises 

Describe  the  surfaces  represented  by  the  following  equations: 
1.  x^  +  y^  =  16.  7.  x«  -  2/2  =  0. 


x^       V^ 
2-   4  +16  =  1- 

3.  a;*  -  t/«  =  1. 

4.  2»  +  y2  =  25. 

5.  22  -  X*  =  25. 
B.  X  +  2y  =  10. 


8.  a;?/  =  1. 

9.  xz  =  2. 

10.  (x  -3)(x  +  2)  =  0. 

11.  y^  =  4x. 

12.  7/2  +  2"  =  2ay. 

13.  x2  +  2/2  =  lOx. 


129.  Partial  Derivatives.  Let  z  =  f(x,  y)  be  a  function  of  two 
independent  variables,  x  and  y.  When  x  takes  on  an  increment 
Ax,  while  y  remains  fixed,  z  takes  on  an  increment  which  we  shall 

denote  by  AxZ.  When  y  takes 
on  an  increment,  Ay,  while  x 
remains  fixed,  z  takes  on  an  in- 
crement which  we  shall  denote 
by  A^2. 

For  example,  if  a  gas  be  en- 
closed in  a  cylinder  with  a  mov- 
able piston,  the  volume  v  of  the 
gas  is  a  function  of  the  tempera- 
ture  T  and  of  the  pressure   p 
which  can  be  varied  by  varying 
the  pressure  on  the  piston.     If 
the  temperature  alone  be  changed  the  volume  will  take  on  a 
certain  increment  A?^.     If  the  pressure  alone  be  changed  the 
volume  will  take  on  the  increment  ApV. 

If  3  =  /(x,  y)  be  represented  by  a  surface.  Fig.  83,  the  increment 
of  z  obtained  by  giving  x  an  increment,  while  y  remains  constant, 
is  the  increment  in  z  measured  to  the  curve  cut  out  by  a  plane 
y  =  k,  Q,  constant.     Thus  A^z  =  HQ;  similarly  A^z  =  KR. 

AxZ 
The  limit  of  the  quotient  -r—  as  Ax  approaches  zero  is  called  the 

partial  derivative  of  z  with  respect  to  x.     It  is  denoted  by  the 

dz 
symbol  -t^.     Then 


dx 


dz 
dx 


lim  AxZ 
^=0  Ax* 


§130]  SOLID  GEOMETRY  247 

It  is  evidently  calculated  from  z  =  J{x,  y)  by  the  ordinary 
rules  of  differentiation,  y  being  treated  as  a  constant.  Thus  if 
z  =  x^y, 

dz 
Geometrically  w—  represents  the  slope  of  the  tangent  at  the 

point  (x,  y,  z)  to  the  curve  cut  from  the  surface  by  the  plane 
through  this  point  parallel  to  the  XZ-plane. 
Similarly 

dz_  _   lim  A^ 

dy       ^J/=o  Ay 

and  it  is  calculated  by  differentiating  z  =  /(x,  y),  treating  x  as  a 
constant.  Geometrically  it  represents  the  slope  of  the  tangent 
at  the  point  {x,  y,  z)  to  the  curve  cut  from  the  surface  by  the 
plane  through  this  point,  parallel  to  the  FZ-plane.     If  z  =  xhj, 

■5—  =  x^. 
dy 

X 

Illustration  1.     If  2  =  sin  -> 

y 


dz  X   d  /x\        1         X 

=  cos  -  -^T"  I  - )   =  -  cos  - 


dx  y  dx  \yj        y        y 

and 

dz  X   d  /x\  X         X 

V-  =  cos  -  ^-  ( - )   = ~  cos  -• 

dy  y  dy  \y}  y^        y 

130.  Partial  Derivatives  of  Higher  Order.  If  z  is  differentiated 
twice  with  respect  to  x,  y  being  treated  as  a  constant,  the  deriva- 
tive obtained  is  called  the  second  partial  derivative  of  z  with  re- 

dh 
spect  to  x.     It  is  denoted  by  the  symbol  3—^.     Similarly  the  second 

partial  derivative  of  z  with  respect  to  y  is  denoted  by  the  symbol 

dy^' 

If  z  is  differentiated  first  with  respect  to  x,  y  being  treated  as  a 
constant,  and  then  with  respect  to  y,  x  being  treated  as  a  constant, 

dh 
the  result  is  denoted  by  the  symbol  „    „  •     If  the  differentiation 

takes  place  in  the  reverse  order  the  result  is  denoted  by  the  symbol 


248  CALCULUS  [§130 

-    „  •     The  first  is  read  "the  second  partial  derivative  of  z  with 
dxatj 

respect  to  x  and  y;"  the  second,  "the  second  partial  derivative 

of  z  with  respect  to  y  and  x."     In  the  case  of  functions  usually 

occurring  in  Physics  and  Engineering,  viz.,  functions  which  are 

continuous  and  which  have  continuous  first  and  second  partial 

d-z  d'^z 

derivatives,     „    ^  ■  =  ^   „  •      The  order     of    differentiation     is 
dyax       oxay 

immaterial. 

Illustration  1.     z  =  xhj. 

dz       ^  d^z       „  dh 

ai  =  2^^^'  d^^  =  2^'  d^x  =  2^- 

^  ^  3.2  ^  =  0  -^-  =  2x 

^2/  '  dy^        '  dxdy 


In  this  case 


d^z  _  d^z 

dydx       dxdy 

X 


IlliLstration  2.   z  = 

dz 
dx  " 

=  sin    • 

y 

1           X 

=  -  cos  — 

y      y 

dx^  " 

1     .     X 

= 2  sin    • 

y'      y 

a?/ax 

1/        .    x\    /      x\        1          X 
=  -  (  —  sin     I  (  —  „ )  —    „  cos  ~ 

y\        yJ  \    yV     2/'      2/ 

= 

1    /      .     X                  x\ 
=    ,  I  a;  sin w  cos  -  )  • 

y'\       y     ^      y) 

dz 

dy  ' 

X            X 

= i  cos  -• 

2/'       y 

d^z 
dy'' 

2x        X       x'   .    X 

=    7  cos iSin  -• 

2/^.         2/       2/^        2/ 

.   dH 
dxdy 

a;    /  .    a;\   /1\        1         x 
=  —i.  I  sm  "  I  I  -  I i  cos 

2/^  \     y)  \yJ     2/'      y 

= 

1     /       .     X                   x\ 
=  —.  I  a;  sin y  cos  - )  • 

2/'  \       y           y) 

Here  again,  we  notice  that 

d'z          dH 
dydx  ~~  dxdy 

§130]  SOLID  GEOMETRY  249 

Exercises 

d^z    d^z      d'^z  d^z 

1.  Find  ^r^»  ^r— ,'  ->    a   '  and  ^   a   '  for  each  of  the  functions : 

(a)  z  =  ^-  (b)  z  =  xy\  (c)  z  =  xhj. 

(d)  z  =  sin  xy.  (e)  z  =  e*  sin  y. 

9^z  6^z 

2.  Find  ,    --  and  -    ^    for  each  of  the  following  functions : 

dydx  dxdy 

(a)  z  =  x^y.  (h)  z  =  xsin'^y.  (c)  z  =  x  cos  y. 

Id)  z  =  y  log  X.  (e)  z  =-  e"  sin  x.  (J)  z  =  y  tan'i  j;. 

It  is  seen  that  ^-^-  =  ^-^T  in  all  of  these  cases. 
dydx       dxdy 

In  the  above  discussion  z  was  considered  to  be  a  function  of  two 
independent  variables  only.  The  notion  of  partial  derivatives 
can,  however,  be  extended  to  functions  of  three  or  more  variables. 

Illustration  3.     li  z  =  xhjt, 

ix  =  2^^^' 

dy      "" '' 
dt  =  ^  ^' 


and 


dy   _^ 
dxdt  "  *'^^' 

d'z      ^2x 
'  dtdydx 


CHAPTER  XV 

SUCCESSIVE  INTEGRATION.     CENTER  OF  GRAVITY. 
MOMENT  OF  INERTIA 

131.  Introduction.  In  the  preceding  chapters  there  have  been 
numerous  examples  of  successive  integration  of  functions  of  a 
single  independent  variable.  Thus,  to  determine  the  law  of 
motion  of  a  falling  body  whose  differential  equation  of  motion  is 

it  is  necessary  to  integrate  twice.    The  result  of  the  first  integration 

ds 
is  -^  =  gt  -{-  Ci,  and  that  of  the  second  is  s  =  ^gt^  +  Cit  +  Cz. 

Exercises 

1.  If  -j—^  =  2x,  find  y  as  a  function  of  x,  given  that  -r-  =  3  when 

X  =  1,  and  y  =  2  when  x  =  4;  given  that  ?/  =  4  when  x  =  2,  and 
that  y  =  7  when  x  =  4. 

d^y 

2.  Find  y  if  -7-^  =  7x.     Assign  suitable  conditions  to  determine  the 

constants  of  integration. 

3.  Find  y  if  ^  =  2x\ 

The  operation  of  finding  the  result  of  Exercise  2  can  be  written 

J[  JiJtx  dx\dx]dz  =  /[/[la;^  +  Ci\dx\dx 

=  ^  +  Ci^-  +  C,x  +  C3. 
The  first  member  can  be  written 

Ix  dx  dx  dx. 


///' 


It  is  a  triple  integral  and  indicates  that  integration  is  to  be  per- 
formed  three   times  in  succession.     An   arbitrary   constant   of 

250 


§132]  SUCCESSIVE  INTEGRATION  251 

integration  is  introduced  with  each  integration.  If  each  integra- 
tion is  performed  between  limits  the  constants  of  integration  do  not 
appear.     Thus, 

7x  dx  dx  dx  =  I     I     -^    dx  dx 


is: 


42  dx  dx 


=  I    42  a;    dx 


■r 


1 


=  1    84  dx 

|2 

=  84  X      =84. 
|i 

132.  Illustration  of  Double  Integration.    Let 

il  =  -'  +  y-  <i) 

Integration  with  respect  to  y,  treating  x  as  a  constant,  gives 

^  =  x^t/  +  |  +  0(x), 

where  (t>(x)  is  an  arbitrary  function  of  x.  This  arbitrary  function 
of  X  takes  the  place  of  an  arbitrary  constant  of  integration  in  the 
case  of  a  single  independent  variable.  A  second  integration,  this 
time  with  respect  to  x,  gives 

2  =  ^  +  ^'  +  J<A(^)rfx  +  ^(^),  (2) 

where  4'{y)  is  an  arbitrary  function  of  y. 

The  result  contains  an  arbitrary  function  of  x  and  an  arbitrary 
function  of  y.  Equation  (2)  represents  a  surface,  but  a  very 
arbitrary  one  on  account  of  the  presence  of  the  arbitrary  functions 
I  (f){x)dx  and  4'(.y)-  The  process  of  finding  (2)  from  (1)  is  indi- 
cated by  a  double  integral  sign.     Thus, 


// 


(x2  +  ?/2)  dy  dx, 


252  CALCULUS  [§132 

which  means 


I    (   I  (x2  +  2/2)di/)  dx. 


Upon  performing  the  integration  indicated,  first  with  respect  to  y, 
then  with  respect  to  x,  we  obtain 

I   I  (x2  +  t/2)  dydx  =    I    (x^y  +J  +  <f>ix))  dx 

Instead  of  an  indefinite  double  integral  such  as  the  one  just 
considered  we  may  have  a  definite  double  integral.  If  the  inte- 
gration with  respect  to  y  is  performed  before  that  with  respect  to 
X,  the  limits  of  integration  with  respect  to  y  may  be  functions  of 
X.    Thus, 

I     \\x^  +  y')  dydx=f  (xH)  +  I)  Y       dx 

The  last  integration  is  readily  j)erformed.  It  is  to  be  noted  that 
in  evaluating  a  double  integral,  x  is  treated  as  a  constant  when  the 
integration  with  respect  to  y  is  performed. 

If  in  a  definite  integral  dx  is  written  before  dy,  the  integration 
with  respect  to  x  is  to  be  performed  first.  ^ 


1.     1      1    xydydx. 

Exercises 

f*ir    f*a{\  +  C03  B) 

4.     1      1                      rdrde. 

Jo  Jo 

2.     1      1      xy  dy  dx. 

Jo  Jo 

r-  rva-^-x^. 

5.    1                      dydx 

Jo   Ja—x 

3.    1      1     xy-'dydx. 
Jo  Jix 

6.     1      1      1    x^y^z^  dz  dy  dx. 
Jo  Jo  Jz 

1  Usage  varies  on  this  point.     The  student  will  have  to  observe  in  every  case  the 
convention  adopted  in  the  book  he  is  reading. 


§133] 


SUCCESSIVE  INTEGRATION 


253 


I  dzdy 

Jo 

nVx  n  rvi 

xdydx.  10.    I      I 

n''  {x^  +  y^)  dy  dx.  11-     I      I 

2  Jo    Jx^ 


dx. 
10. 


12. 


dy  dx. 


y  dy  dx. 


Vo^  —  ^"^  —  y^  dy  dx. 


Hint.  To  perform  the  integration  in  Exercise  12,  let  v  a^  —  x^  =  b 
and  make  use  of  the  result  of  Illustration  3,  §105. 

133.  Area  by  Double  Integration:  Rectangular  Coordinates. 
A  plane  area  can  be  represented  by  a  double  integral.  Thus,  let 
it  be  required  to  find  the  area  A  between  the  curves  y  =  /i(x), 
y  =  f^ix),  and  the  lines  x  =  a  and  x  =  b.  The  area  of  the  strip 
IJKH,  Fig.  84,  is  approximately 


lim 

Ay 


«2  «t  /.V, 

j2)At/Aa;  =  Aa;]^m^2)Ay  =  Ax  I     dy, 


where  yi  and  y^  are  the  ordinates  of  the  two  curves  y  =  /i(x)  and 
y  =  fiix),  respectively.    And 
the  area  sought  is  approxi- 
mately 


x  =  a       Jvi 


Fig.  84. 


The  smaller  Ax  is  taken,  the     "o 
closer    the     approximation. 
The  limit  of  this  sum  as  Ax 

approaches  zero  is  the  area  sought.     Since  y\  and  2/2  are  func- 
tions of  X,  1      dy  is,  a,  function  of   x  and  consequently 


lim 


K)X   Ax   I     ^2/=    I       \    dy 


dx  =  A. 
It  is  to  be  noted  that  in  setting  up  this  integral  the  summation 


254 


CALCULUS 


[§133 


with  respect  to  y  was  performed  first,  giving  the  area  of  a  vertical 
strip  for  a  particular  value  of  x.  Consequently,  the  integration 
with  respect  to  y  is  to  be  performed  first,  x  being  treated  as  a 
constant.  On  performing  the  integration  with  respect  to  y  we 
obtain 

(2/2  -  yi)dx  =    I    [/zCx)  -  Si{x)\dx, 

a  Ja 

a  single  integral  which  might  have  been  set  up  at  once  by  consider- 
ing the  area  as  the  sum  of  vertical  strips  of  length  1/2  —  yi,  and 
of  width  Ax.  It  is,  however,  desirable  to  be  able  to  set  up  a 
double  integral  over  an  area. 

In  choosing  the  limits  for  a  double  integral,  the  student  should 
proceed  systematically.     The  process  of  setting  up  the  above 

integral  with  its  limits  is  the  follow- 
ing: The  "element"  is  the  rectan- 
gular element  of  area  dy  dx.  The 
"summation"  (integration)  of  this 
element,  for  a  particular  value  of  x, 
between  the  limits  for  y  of  WI  and 
WH,  the  ordinates  of  the  curves  y  = 
/i(x)  and  y  =  /2(.t),  gives  the  area  of 
the  typical  strip  IJKH.  The  "sum- 
mation" (integration)  of  the  strips  of 
which  this  is  a  typical  one,  between 
the  extreme  values  oi  x,  x  =  a  and  x  =  b,  gives  the  area  sought. 
Thus 

A  =   I     I     dy  dx. 


nv 


The  procedure  may  be  briefly  summarized  in  the  following  concise 
directions.  Write  first  the  element  dy  dx,  then  the  integral  sign, 
then  the  limits  /i(x),  /2(x),  then  another  integral  sign  with  the 
limits  a  and  b. 

Illustration.  Find  by  double  integration  the  area  between  the 
parabolas  y^  =  x  and  y  =  x^.  The  integral  is  set  up  as  follows: 
Write  the  element  dy  dx,  then  an  integral  sign  with  the  limits 
x^  and  \/x.  This  represents  the  area  of  the  typical  strip,  IJKH, 
Fig.  85,  for  a  fixed  x.    All  of  the  strips  of  which  this  is  a  typical 


§134]  SUCCESSIVE  INTEGRATION  255 

one  are  to  be  summed  from  a;  =  0  to  x  =  1,  the  abscissas  of  the 
points  of  intersection  of  the  curves.  Then  write  the  second 
integral  sign  preceding  the  first  with  the  limits  0  and  1.     Thus 


Jo   Jx' 


dy  dx  =  \. 


Exercises 

1.  Find  by  double  integration  the  area  between  the  curves  y  =  x 
and  y^  =  x'. 

2.  Find  the  area  of  Exercise  1  by  integrating  first  with  respect  to  x 
and  then  with  respect  to  y. 

3.  Find  by  double  integration  the  area  between  r/*  =  a{a  —  x)  and 
X*  +  2/^  =  a^. 

4.  Find  the  area  between  y^  =  ax  and  y^  =  2ax  —  x^. 

5.  Find  the  area  of  Exercises  3  and  4  by  integrating  first  with  respect 
to  X. 

6.  Find  the  area  bounded  by  y^  =  4x,  x  +  y  =  3,  and  the  X-axis. 

7.  Find  the  area  of  a  rectangle  by  double  integration. 

8.  Find  the  smaller  area  between  x^  +  y^  =  1  and  y  =  x  +  5- 
134.  Geometrical  Meaning  of  the  Definite  Double  Integral 

Consider  the  definite  double  integral 

fix,  y)  dy  dx.  (1) 

,  =  /i(x) 


J'»6         /»2/2=/20 


In  accordance  with  the  definition  of  a  definite  single  integral, 
§66,  (1)  can  be  written 

f'[ll^oXf(^'y)  ^y]d^'  (2) 


Here  x  is  considered  constant  under  the  summation  sign,  and 
f{x,  y)  is,  for  such  a  fixed  x,  a  function  of  y  alone. 

V, 

lim 


^2  /»2/2 


is  a  function  of  x,  since  x  occurs  as  an  argument  of  /  and  also  in 
the  limits  of  integration.     Hence  we  can  write  (2)  in  the  form 


256 


CALCULUS 


[§134 


lim 


O  L  «.  J 


-E%'Zll%Xn-.y)^vAx, 


(3) 


where  Ax  under  the  second  summation  sign  is  regarded  as  a  con- 
stant multiplier. 

Z 


Fig.  86. 

In  Fig.  86,  let  EFGL  represent  the  surface  z  =  f{x,  y) ;  QABS, 
in  the  ZF-plane,  the  curve  y  =  fi{x) ;  DHKC,  the  curve  y  =  fi{x) ; 
AD  the  line  x  =  a;  BC  the  line  x  =  b;  and  A'B'C'D'  the  portion 
of  the  surface  cut  from  z  =  f(x,  y)  by  the  cylinders  y  =  fi{x), 
y  =  f^ix),  and  the  planes  x  =  a  and  x  =  b. 


§134]  SUCCESSIVE  INTEGRATION  257 

Divide  A  BCD  into  small  rectangles,  as  shown  in  the  figure,  by 
lines  parallel  to  the  X-  and  F-axes,  at  intervals  of  At/  and  Ax, 
respectively.  Through  these  lines  pass  planes  parallel  to  the 
XZ-  and  FZ-planes.  These  planes  divide  the  solid  bounded  by  the 
planes  and  surfaces  of  Fig.  86  into  vertical  columns  of  rectangular 
cross  section  Ay  Ax.  The  column  erected  on  MNPR  as  a  base  is  a 
typical  one.  f{x,  y)Ay  Ax  represents  approximately  the  volume  of 
the  column  whose  base  is  MNPR  and  whose  top  is  M'N'P'R',  since 
the  area  of  its  base  is  Ay  Ax  and  its  altitude  is  MM'  =  f{x,  y). 
Then  the  sum  of  the  columns  at  a  fixed  distance  from  the 
yZ-plane, 

2)/(x,  y)  Ay  Ax, 

is  approximately  the  volume   of  the  slab  between  the  planes 
IHH'I'  and  JKK'J',  i.e.,  between  the  planes  x  =  x  and 
X  =  X  +  A  X.     And 

h 


0  r  «2  -1 

a  L  w.  -I 


the  sum  of  the  volumes  of  all  the  slabs,  is  approximately  the  vol- 
ume of  the  solid  ABCDA'B'C'D'.  If  Ay  and  Ax  are  each  taken 
smaller  and  smaller  this  sum  will  eventually  represent  a  very  close 
approximation  to  the  volume  in  question,  and  the  limit  of  this 
sum  as  Ay  and  Ax  approach  zero  is  the  volume.  Hence  the  inte- 
gral 

f(x,  y)  dy  dx, 


which  we  have  seen  is  equal  to 

X  =  b  r-  V 


a;  =  a         L  v.  J 


lim 

x  =  a 


represents  the  volume  bounded  by  the  plane  z  =  0,  the  surface 
^  =  /(-^j  y),  the  planes  x  =  a  and  x  =  h,  and  the  cylinders 
y  =  /i(.t)  and  y  =  Jiix). 

Illustration.     Find  the  volume  contained  in  the  first  octant  of 

17 


258 


CALCULUS 


[§135 


the  sphere  x^  +  y^  +  z"^  =-  a^.     See  Fig.  87.     The  equation  of  the 
surface  is 

z  =  \/a2  —  x^  —  y^. 

yi  =  fiix)  =  0 

2/2  =  f2{x)  =  Va^  -  x\ 

the  trace  of  the  sphere  on  the  XF-plane.     The  volume  of  the  col- 
umn on  MNPR  as  a  base  is 

Va^  —  x^  —  2/2  A?/  Ax, 

or,  as  we  shall  say  in  the  future, 

\^a^  —  x^  —  y^  dy  dx. 

The  summation   of  these  columns 
for  a  fixed  x  gives 


Fig.  87. 


r 


■y/a'^  —  x^  —  y"^  dy  dx, 


!/■ 


the  volume,  expressed  as  a  function  of  x,  of  the  slab  between  the 
planes  x  =  x  and  x  =  x4-Axorx  =  x  +  dx.  The  summation 
of  all  these  slabs  from  x  =  0  to  x  =  a  gives 

*y/a^  —  xi       

^/a^  —  x^  —  y^  dy  dx, 
to 

the  volume  of  one  octant  of  the  sphere.  This  integral  was  evalu- 
ated in  Exercise  12,  §132. 

Exercises 

1.  Find  the  volume  of  the  segment  of  the  paraboloid  y^  +  2z^  =  4x, 
cut  off  by  the  plane  x  =  5. 

2.  Find  the  volume  bounded  by  the  cylinders  y  =  x^  and  y-  =  x, 
and  the  planes  z  =  0  and  z  =  1. 

3.  Find  the  volume  common  to  the  cylinders  x^  +  y^  =  a^  and 
y^  +  z^  =  a2. 

4.  Find  the  volume  between  the  cylindrical  surface  y^  —  x',  the 
plane  y  =  x,  and  the  planes  z  =  0  and  z  =  1. 

135.  Area:  Polar  Coordinates.    Let  it  be  required  to  find,  by 
double  integration,  the  area  between  the  radii  vectores  6  =  a, 


§135] 


SUCCESSIVE  INTEGRATION 


259 


and  d  =  ^,  and  the  curve  p  =  f{d).  Divide  the  area  as  shown  in 
Fig.  88,  the  radii  making  an  angle  of  Ad  with  each  other  and  the 
radii  of  the  concentric  circles  differing  by  Ap.  The  area  of  MPQR 
is  equal  to 

Kp  +  Ap)2  A0  -  §  p2  A^  =  p  Ap  A0  +  ^  Ap2  A0 

As  Ap  approaches  zero, 

limpApA0  +  |Ap2A5 


Hence 

P=Ke) 


^"=0  p  Ap  A^ 


=  1. 


p  =  0  p  =  0  Jp  =  0 


Fig.  88. 


Fig.  89. 


This  sum  represents  the  area  of  the  sector  OHL.  The  total  area 
sought  is  the  limit  of  the  sum  of  these  sectors  as  A9  approaches 
zero,  i.e. 

^  =  Ai"oX^^        p^p=  \       p^p^- 


This  integral  is  to  be  set  up  as  follows:  The  element  of  area  is 
the  approximately  rectangular  area  MPQR  whose  area  is  approxi- 
mately {MR){MP)  =  pdpdd.  This  element  is  to  be  summed 
from  p  =  0  to  p  =  f{d) .  This  gives  approximately  the  area  of  the 
typical  sector  OHK.  These  sectors  are  to  be  summed  from 
e  =  atod  =  fi. 


260  CALCULUS  [§136 

More  briefly:  Write  down  the  element  pdpdd,  then  an 
integral  sign.  Its  limits  are  the  extreme  values  of  p  for  a  given  6. 
Then  write  before  this  integral  another  integral  sign.  Its  limits 
are  to  be  chosen  so  as  to  sum  up  all  the  sectors  such  as  OHK. 

Illustration  1.  Find  the  area 
of  the  circle  p  =  10  cos  6,  Fig. 
89.  The  area  bounded  by  the 
semicircle  above  the  initial  line 
will  be  found  and  multiplied  by 
two. 


.20   cos  ff 


Jo    Jo 


lOcosO 

A  =2\      I         pdpde. 


p  dd  dp. 

Illiistration  2.     Find  by  double  integration  the  area  between 
p  =  10  cos  e  and  p  =  20  cos  d.     See  Fig.  90. 


•20  COS  e 
A  =2\      I  pdpdd 

'l0CO3  tf 

Show  that 


t/O      t/10 

mo    /•«=o«-'^  /•20     /-cos-^ 

A  =  2  I       I        ^    pdddp+2  \       I  pdddp. 

Jo       JC08-1-^  JlO       Jo 


10 

Which  method  is  the  simpler  in  this  case? 

Exercises 

Find  by  double  integration: 

1.  The  area  of  the  circle  p  =  5  sin  0. 

2.  The  area  of  the  cardioid  p  =  a(l  —  cos  9). 

3.  The  area  of  the  lemniscate  p"  =  a"  cos  20. 

4.  The  area  outside  p  =  a  (1  +  cos  6)  and  inside  p  =  3a  cos  d. 
136.  Volume  of  a  Solid:  Triple  Integration.     We  shall  now 

find  the  volume  of  the  solid  of  Fig.  86  by  triple  integration.     Sup- 


§136]  SUCCESSIVE  INTEGRATION  261 

pose  the  solid  further  subdivided  by  planes  parallel  to  the  XY- 
plane  and  at  a  distance  Az  apart,  into  rectangular  parallelopipeds 
of  Volume  Az  Ay  Ax.  Then  the  volume  of  the  column  on  the  base 
MNPR  is  approximately 

a  =  f(.x,  y) 

^      Az  Ay  Ax. 

Z   =    0 

Then 

Vi  z  =>  f(.x,y) 

'^       2^      Az  Ay  Ax 

y\       «  =  0 

is  approximately  the  volume  of  the  slab  between  the  planes 
IHI'H'  and  JKJ'K',  i.e.,  between  the  planes  x  =  x  and 
X  =  X  ■\-  Ax. 

X  =  h    2/2     2  =  f{x,  y) 

X    2)       2)      Az  Ay  Ax 

X  =  a    y^        «  =  0 

is  approximately  the  sum  of  the  volumes  of  all  of  the  slabs. 
If  Az,  Ay,  and  Ax  are  each  taken  smaller  and  smaller,  this  sum  will 
represent  a  very  close  approximation  to  the  volume  sought,  and  the 
limit  of  this  sum  as  Ao:,  Ay  and  Az  approach  zero  is  exactly  this 
volume.     Hence  the  integral. 


a    Jy^    Jo 


dz  dy  dx, 


represents  the  volume  bounded  by  the  plane  z  =  0,  the  surface 
z  —j{x,y),  the  planes  x  =  a  and  x  =  h,  and  the  cylinders 
y  =  j\{x)  and  y  =  Si{x). 

Illustration  1.     Find  by  triple  integration  the  volume  of  the 
ellipsoid 

a"  ^  62  -r  g2 
See  Fig.  91. 

F  =  8   I       I  dzdydx 

Jo     Jo  Jo 


262 


CALCULUS 


[§136 


The  student  will  perform  the  integration. 

Illustration  2.  Find  by  triple  integration  the  volume  of  the 
solid  bounded  by  the  cylinder  x^  -\-  y^  =  2ax,  the  plane  z  =  0, 
and  the  paraboloid  of  revolution  x^  -{-  y^  =  ^az.  Write  the  ele- 
ment of  volume,  dz  dy  dx.    The  integration  with  respect  to  z 

cc^  "4*  1/^ 
-J gives  the  volume  of  the  typical 


between  the  limits  0  and 


vertical  column  of  base  dy  dx,  extending  from  the  point  (x,  y)  in 

/j;2   _|_    y2 

the  plane  z  =  0  to  the  surface  of  the  paraboloid,  z  =  — j ' 

Next,  X  being  kept  fixed,   these  columns  are  summed  into  a 
typical  slab  by  integrating  with  respect  to  y  from  the  Z-axis, 


y  =  0,  to  y  =  ^/2ax  —  x^,  the  trace  of  the  cylinder  in  the  XY- 
plane.  Finally  the  integration  with  respect  to  x  from  x  =  0  to 
X  =  2a  gives  one-half  of  the  total  volume  sought,  viz.,  that  lying 
in  the  first  octant. 

xi  +  y^ 

F  =  2  I         I  I  dzdy  dx. 


Jo      Jo  Jo 

The  student  will  perform  the  integration. 


Exercises 

1.  Find  the  volume  common  to  the  cylinders  x"  +  y^  =  r^  and 

x2  -f-  22  =  r^ 


§137]  SUCCESSIVE  INTEGRATION  263 

2.  Find  the  volume  of  one  of  the  wedges  cut  from  the  cylinder 
^2  ^  y2  —  J.2  \yy  ^hc  plancs  z  =  0  and  z  =  mx. 

3.  Find  the  volume  in  the  first  octant  bounded  by  the  coordinate 

X       y       z 
planes  and  the  plane  — h  r  H —  =  1. 
^  ^         a      0       c 

4.  Set  up  the  integral  representing  the  volume  bounded  by  the 
surface  x^  -\-  y^  +  z^  =  o*. 

6.  Find  the  volume  between  y^  +  z^  =  4ax  and  x  —  z  =  a. 

6.  Find  the  volume  between  the  planes  y  =  0,  z  =  0  and  the  sur- 
faces z  =  x^  +  Ay^,  y  =  1  —  x'^. 

7.  Find  the  volume  between  y^  +  2z^  =  4x  and  z  =  x. 

137.  Center  of  Mass,  Centroid.  Let  there  be  a  system  of 
masses  mi,  m2,  ms,  .  .  .  ,  rUn  situated  at  the  points  {xi,  yi,  Zi), 
{xi,  2/2, 22),  (^3, 2/3,  Zz),  .  .  .  ,  {Xn,  yn,  Zn) ,  Tespectlvely.  The  mean 
distance  with  respect  to  mass,  of  the  system  from  the  FZ-pIane  is 

The  mean  distances,  with  respect  to  mass,  of  the  system  from  the 
ZX-  and  XF-planes  are,  respectively, 

2  m.i/< 


and 


(2) 


(3) 


The  point  (x,  y,  z)  is  called  the  centroid,  or  the  center  of  mass,  of 
the  system  of  masses  mi,  m2,  •    •    • ,  m„. 

m,-  Xi  is  called  the  moment,  and  Xi  the  moment  arm,  of  the  mass 
mi  with  respect  to  the  FZ-plane.^  Then  x  is  the  mean  moment 
arm  with  respect  to  the  FZ-plane  of  the  masses  mi,  m2,  •  •  •,  m„. 
For,  equation  (1)  shows  that  if  all  the  masses  were  placed  at  the 
distance,  x,  from  the  FZ-plane,  the  moment  with  respect  to  this 
plane  would  be  the  same  as  the  sum  of  the  moments  of  the  masses. 
Hence  we  can  say  that  the  centroid  of  a  system  of  masses  is  a  point 
such  that  if  all  the  masses  were  concentrated  at  this  point,  the 
moment  with  respect  to  each  coordinate  plane  would  be  equal  to 

1  The  term  moment  of  a  mass  with  respect  to  a  plane  has  evidently  a  diflerent 
significance  from  the  term  moment  as  applied  to  a  force. 


264  CALCULUS  [§139 

the  sum  of  the  moments  with  respect  to  the  corresponding  planes 
of  the  masses  in  their  given  positions. 

138.  Centroid  Independent  of  the  Position  of  the  Coordinate 
Planes.  It  will  now  be  shown  that  the  distance  of  the  centroid, 
(x,  y,  z),  from  any  plane  is  the  mean  of  the  distances,  with  respect 
to  mass,  of  the  masses  mi,  m2,  •  •  •,  m„,  from  that  plane.  And 
thus  it  will  be  shown  that  (x,  y,  z),  the  centroid,  is  a  point  whose 
position  with  reference  to  the  masses  is  independent  of  the  choice 
of  the  coordinate  planes. 

Let  ax  -\-  hy  -\-  cz  -\-  d  =  0  hQ  the  equation  of  a  given  plane 
(see  §120).  The  distance,  p,  of  the  point  (x,  y,  z)  from  this  plane 
is 

-      ax  +  hy  +  cz  ■\-  d 

P  = 22 '  ^^^ 

where  

R  =  Va^  +  62  +  c\ 

(See  §124.)  On  substituting  the  values  of  x,  y,  and  2  from  (1), 
(2),  and  (3)  §137,  and  reducing  the  absolute  term,  d,  to  the  common 
denominator,  we  have 

aSwiX,-  +  blfTTiiyi  +  cZniiZi  +  dXnii 


or 


sm.r 


R'Emi 
axi  +  hyi  +  czi  +  d' 


R  J.  (2) 

ox,-  +  hyi  •\-czi  +  d  •      +1       j-  .  f     .1  •  ^ 

But   5 =  pi    is    the    distance    of     the    point 

K 

(x,-,  yi,  Zi)   from  the  given  plane.     Hence   (2)   can  be  written 

in  the  form 

_      S  mipi 

This  proves  the  statement  at  the  beginning  of  this  section.  In 
other  words,  if  all  the  masses  of  the  system  were  concentrated  at 
the  centroid,  the  moment  with  respect  to  any  plane  would  be 
equal  to  the  moment  of  the  system  of  masses  with  respect  to  this 
plane. 

139.  Center  of  Gravity.  Let  the  system  of  masses  considered 
above  be  acted  upon  by  gravity.     It  will  be  shown  that  the  line 


§140]  SUCCESSIVE  INTEGRATION  265 

of  action  of  the  resultant  force  passes  through  the  center  of  mass, 
or  the  centroid. 

Since  the  position  of  the  centroid  is  independent  of  the  choice  of 
axes,  choose  the  positive  direction  of  the  axis  of  z  vertically  upward 
and  the  axes  of  x  and  y  horizontal.  The  force  acting  on  mi  is 
mi^,  that  on  m^  is  m-ig,  etc.  The  resultant  force  is  equal  to  2  niig 
and  is  directed  vertically  downward.  Its  line  of  action  meets  the 
XF-plane  in  a  point  (a,  j3,  0)  such  that  its  moment,  aSm.gr,  about 
the  F-axis  is  equal  to  2m,gfXi,  the  sum  of  the  moments  of  the 
forces  acting  on  the  individual  masses;  and  such  that  the  moment, 
jSSmigr,  about  the  X-axis,  is  equal  to  the  sum  of  the  moments, 
Sw.grr/,,  of  the  forces  about  this  axis. 


and 
Whence 

and 

Consequently 


Hence  the  resultant  passes  through  the  centroid  of  the  system  of 
masses. 

If  the  masses  all  lie  in  one  plane,  say  the  XF-plane,  z  is  zero  and 
the  centroid  is  fixed  by  the  two  coordinates  z  and  y.  The  product 
rtiiXi  is  called  the  moment  of  the  mass  rrii  with  respect  to  the 
F-axis.  In  this  case  x  is  the  mean  moment  arm  with  respect  to 
the  F-axis. 

If  the  masses  all  lie  upon  a  line,  say  the  X-axis,  the  centroid  is 
fixed  by  a  single  coordinate,  x. 

140.  Centroid  of  a  Continuous  Mass.  If  instead  of  discrete 
masses  we  have  a  continuous  mass,  the  coordinates  of  the  center 
of  mass,  or  the  centroid,  are  clearly. 


a2,m 

iQ  =  'Lrrii 

g^i, 

/SSw 

xQ  =  'Lrui 

gvi- 

^rUiXi 

Oi  = 

^rrii 

^  = 

^rmyi 

2)^< 

a  = 

X  and  |8 

=  y- 

266  CALCULUS  [§140 


Ani  = 


X  = 


y  = 


i%^Am         Jdm 


z  = 


IL^.oX^         /dm 
^jj  Vz  Am        J  z  dm 
^  X  -^m        fdm 


lim 

Am 


Am=( 

The  integration  is  to  be  extended  throughout  the  entire  mass,  and 
the  integrals  considered  may  be  single,  double,  or  triple,  depending 
on  the  form  of  the  mass. 

Illustration  1.  Find  the  center  of  gravity  of  a  bar,  Fig.  92, 
of  length  L,  whose  linear  density,  p,  may  vary.  Let  the  axis  of  x 
coincide  with  the  bar,  the  origin  being  taken  at  one  end.    The 


OF 


^1 


Fig.  92. 

mass  of  an  "element"  of  the  bar  of  length  dx  is  p  dx,  p  being  a 
function  of  x,  the  distance  of  the  element  from  the  origin.  The 
moment  of  this  element  of  mass,  dm  =  p  dx,  about  an  axis 
through  the  origin  perpendicular  to  the  bar  is 

X  dm  =  xp  dx. 

X,  the  abscissa  of  the  centroid,  the  only  coordinate  necessary 
to  fix  the  centroid  in  this  case,  is  given  by 

px  dx 

mJ  0 

X  = 


f. 


p  dx 


The  numerator  represents  the  total  moment,  and  the  denominator 
the  total  mass.     If  the  bar  is  of  uniform  density,  p  can  be  taken 


il40] 


SUCCESSIVE  INTEGRATION 


267 


out  from  under  the  integral  sign.     Then 

xdx      — 
2 

X  = 


r 


^L. 


dx 


x\ 


If  the  linear  density  is  proportional  to  the  distance  from  one  end, 
then  p  =  kx  and  we  have 

k   \     x^  dx        1 


f 


k 


f. 


X  dx 


Illustration  2.  Let  it  be  required  to  find  the  center  of  gravity 
of  a  plate  of  uniform  thickness  and  of  mass  p  per  unit  volume  or  of 
mass  p  per  unit  surface.  Take  a  plate  of  the  shape  of  Fig.  84. 
The  mass  of  the  element  MNPR  is  p  dy  dx.  The  moment  of  this 
element  about  the  Y-  axis  is  xp  dy  dx,  and  its  moment  about  the 
X-axis  is  jj  p  dy  dx.     Then 


r  r 


p  X  dy  dx 


fdm  r    p  p^y^^ 


p  y  dy  dx 


If  p  is  constant, 


and 


i  ^^«  r  r^pdydx 

Ja     Jvi 

I      xdy  dx 

I  {      dy  dx 

Ja     Jy^ 

J'*b    nvi 
I      ydydx 
a    Jy^ 

y^    rb  /»!/,  ' 

II  dy  dx 

Ja  Jv^ 


268  CALCULUS  [§140 

The  numerator  of  each  of  these  expressions  is  the  integral  of  the 
product  of  an  element  of  area  by  its  distance,  x  or  y,  from  the  Y- 
axis  or  X-axis,  respectively.  The  denominator  is  the  area.  The 
mass  does  not  enter  into  either  of  these  formulas.  We  are  thus 
led  to  speak  of  the  centroid  of  an  area,  of  a  line  or  of  a  solid,  with- 
out reference  to  its  mass.  This  notion  of  the  centroid  of  a  geo- 
metrical figure,  a  line,  an  area,  or  a  solid,  without  reference  to  its 
material  composition  is  an  important  one.  For,  in  many  prob- 
lems in  mechanics  one  is  interested  in  the  centroid  of  a  geomet- 
rical configuration  as  such.  Thus  in  the  study  of  the  deflection 
of  beams  it  is  necessary  to  know  the  position  of  the  centroid  of  the 
cross  section  of  the  beam. 

Illustration  3.  Find  the  centroid  of  the  solid  represented  in 
Fig.  86.  The  element  of  mass,  dm,  is  equal  to  p  dz  dy  dx,  and  its 
moment  with  respect  to  the  yZ-plane  is  xp  dz  dy  dx.     Then 

Vix,  y) 

px  dz  dy  dx 


I  X  dm 
I  dm 


Ja    Jy^     Jo 


Similarly : 


Ja   Jy^    Jo 
J     Jo 


'fix,  y) 

p  dz  dy  dx 


'/(x,  y) 

py  dz  dy  dx 


and 


_        \y  dm 

J^^  I  Pdzdydx 

Ja    Jvi     Jo 


j  zdm 
1  dm 


rb    ny,    n 

Ja    Jvi     Jo 


'Kx,  v) 

pz  dz  dy  dx 


rb    ry,    r 

Ja    Jy^     Jo 


iix,  y) 

p  dz  dy  dx 


If  the  density  is  constant,  p  can  be  canceled  from  numerator 
and  denominator. 

If  the  solid  has  an  axis  or  a  plane  of  symmetry  the  centroid  lies 
in  this  axis  or  in  this  plane. 

Illustration  4.  Find  the  centroid  of  the  area  in  the  first  quad- 
rant bounded  by  the  circle  x^  -\-  y^  =  a^. 


§140]  SUCCESSIVE  INTEGRATION  269 

If  we  use  double  integration  we  have,  in  accordance  with  Illus- 
tration 2, 

X  dy  dx 


n 

—       Jo   Jo 


4 
and 

ydydx 

y  =  — "—^ 

4 
Radicals  could  be  avoided  in  the  evaluation  of  the  numerator 
of  the  expression  for  x  if  the  integration  were  performed  first  with 
respect  to  x  and  then  with  respect  to  y.    Thus 

"Va2  -  j/2 

X  dx  dy 

'0    «/o 
X  =   ^ • 

1" 

The  student  will  evaluate  each  expression  given  for  x. 

From  the  symmetry  of  the  figure,  x  =  y,  and  it  is  not  necessary 
to  evaluate  the  integral  for  y. 

In  finding  the  centroid  in  this  case,  and  indeed  in  many  cases, 
it  is  easier  to  use  single  integration  than  double  integration. 
Thus  if  we  choose  as  the  element  of  area,  the  strip  y  dx  parallel 
to  the  y-axis,  the  moment  of  this  strip  about  the  F-axis  is  xy  dx, 
and 

X  dm        I    xy  dx         I    x\/a'  —  x^  dx 
Jo  Jo 


h     f 


ydx  -^ 


Illustration  5.     Find  the  centroid  of  the  solid  in  the  first  octant 
bounded  by  the  sphere  x^  -\-  y^  -\-  z^  =  a^. 
The  method  of  Illustration  3  gives 

I  X  dz  dy  dx 

Jo  Jo  Jo  ' 

X  =  -, 


270  CALCULUS  (§140 

From  considerations  of  symmetry,  y  =  z  =  x. 

Here  again  it  is  simpler  to  use  single  integration.  Choose  as 
element  a  slab  of  thickness  dx  parallel  to  the  FZ-plane.  The 
base  of  such  a  slab  is  a  quadrant  of  a  circle  of  radius  \/a^  —  x^, 
where  x  is  the  distance  of  the  slab  from  the  FZ-plane.  The  volume 
of  this  elementary  slab  is 


7r(a^  —  x^)   , 
z ax. 


Hence 

TT 


X   = 


I    X  {a^ 


.  ,    .^  .  V.        x^)  dx 
4.  f       ^ 


Exercises 

Find  the  coordinates  of  the  centroid  of: 

1.  The  area  between  y  =  x^  and  y^  =  x, 

2.  The  areas  of  Exercises  1,  3,  and  6,  §133. 

3.  A  triangular  plate. 

Hint.     Draw  lines  parallel  to  the  base,  BC,  Fig.  93,  at  intervals 
dx  along  the  median  AM.     The  mass  of  each  strip  is  proportional 

to  AL  =  X  and  can  be  regarded  as 
concentrated  at  its  centroid  on  the 
line  AM.  Hence  we  can  think  of 
the  triangular  plate  as  replaced  by 
the  bar  AM  whose  density  is  propor- 
tional to  the  distance  from  the  end 
A  In  accordance  with  Illustration 
1,  its  centroid  is  at  a  point  /  two- 
FiG.  93.  thirds  of  the  way  from  A  to  M. 

The  centroid  of  a  triangle  can  also 
be  located  without  any  calculation  whatever.  From  Fig.  93  it  follows 
that  the  centroid  lies  on  the  median  AM.  The  same  argument  shows 
that  it  lies  on  the  medians  BMi  and  CM2.  Hence  it  lies  at  the 
point  of  intersection  of  the  medians,  i.e.,  at  a  point  two-thirds  of  the 
way  from  a  vertex  to  the  middle  of  the  opposite  side. 

4.  The  area  of   a  semicircular  plate  of  radius  r.     (Single  integra- 
tion will  be  sufficient.) 


jl40] 


SUCCESSIVE  INTEGRATION 


271 


5.1  Let  OMKB,  Fig.  94,  be  a  quadrant  of  a  circle  of  radius  r.  Let 
OMDB  be  a  square.  Denote  by  Ci,  C2,  and  C3  the  centers  of  gravity 
of  the  square,  the  quadrant  of  the  circle,  and  the  area  MDBKM, 
respectively;  and  by  Ai,  A2,  and  A3  the  corresponding  areas.     Then 

^2X2  +  AzX3  =  AiXi 

A2X2 


AiXi 
X3  = 


=  0.223  r. 


DC3  =  0.315  r. 
6.  A  circular  arc  of  radius  r  and  central  angle  2a.     See  Fig.  95. 


0 

Ai 

\ 

/ 

/ 

\ 

\"~ 

/C2 

*-     \ 

V 

/ 

-^^ 

D 

t 

i 

Fig.  94. 


Fig.  95. 


Hint.  The  centroid  lies  on  the  radius  which  bisects  the  central 
angle  since  this  line  is  an  axis  of  symmetry.  Choose  this  radius  as  the 
axis  of  X  and  the  center  of  the  circle  as  the  origin.     Then  y  =  0,  and 


-  Xr!  X 

^  ~        Ira.       ~ 


r  cos  Or  do      r'- 

—  a 

'  Ira. 


/:.• 


cos  0  do 


2rc 


This  problem  can  also  be  solved  by  using  rectangular  coordinates. 
Thus 


—  t/r  C( 


xdx 

Wj.2  — 


2rc 


2r^  sin  a       r  sin  a 
2ra 


7.  The  portion  of  the  arc  of  the  circle  x^  +  j/^  =  r^  which  lies  in  the 

1  Exercises  5,  9,  20,  21,  and  22  taken  from  Technical  Mechanics  by  Maurer. 


272 


CALCULUS 


[§140 


first  quadrant.     Use  the  result  of  Exercise  6.     Also  find  the  result 
directly. 

8.  The  parabolic  segment  of  altitude  a  and  base  h.     See  Fig.  96. 
Hint.     Show  that  the  equation  of  the  parabola  is  ^ay"^  =  b^x. 

Ans.  X  =  fa. 

9.  A  conical  or  pyramidal  solid  of  altitude  a  and  base  A. 

Hint.  Let  OMNO,  Fig.  97,  represent  the  projection  of  the  solid 
on  the  XF-plane.  Divide  the  solid  by  planes  parallel  to  the  base 
into  laminas  or  plates  of  thickness  dx.     Then  the  area  of  the  lamina 


' 

Y 

N 

^.---^ 

~" 

" 

< 

a >^> 

o 

f 

a 

'  K 

^ 

/ 

/^ 

\ 

^^...^M 

\ 

^--^^ 

\ 

, 

O 

^ 

•"■""'''^      !    !                    X 

dx 

Fig.  96. 

Fig. 

97 

, 

Ax"^ 


whose  abscissa  is  x  is  — ^:  and  its  volume  is 
solid  is  -0-.     Hence : 


Ax^dx 


The  volume  of  the 


J'*''     /Ax^dx\ 
X  = 1 


Aa 
T 


3a 
4" 


Further  the  centroid  of  every  lamina  lies  on  the  line  joining  the  apex 
with  the  centroid  of  the  base.  Consequently  the  centroid  of  the 
solid  lies  on  that  line. 

10.  The    hemisphere    generated    by    revolving    one   quadrant   of 
x»  -(-  j/2  =  r^  about  the  A'-axis.     Evidently  7/  =  2  =  0  and 


■I 


xy'  dx 


11.  The  surface  of  the  hemisphere  of  Exercise  10. 

12.  The  segment  of  a  paraboloid  of  revolution  of  altitude  h. 


§140] 


SUCCESSIVE  INTEGRATION 


273 


13.  The  semi-ellipsoid  of  revolution  generated  by  revolving  one 
quadrant  of  -^  +  r^  =  1  about  the  X-axis. 

14.  The  surface  of  the  paraboloid  of  Exercise  12. 

15.  The  surface  of  a  right  circular  cone.     Any  conical  or  pyramidal 
surface. 

16.  The  area  of  the  cycloid  a;  =  a{d  —  sin  e),y  =  a(l  —  cos  6). 

17.  The  arc  of  the  cycloid  of  Exercise  16. 

18.  The  area  in  the  first  quadrant  under  x^  +  y^  =  a^. 

19.  The  arc  of  the  curve  of  Exercise  18  in  the  first  quadrant. 

20.  The  segments  of  the  ellipse  indicated  in  Fig.  98.     It  will  be 
found  that  the  centroid  of  the  segment  XAAX  coincides  with  that  of 


Fig.  98. 


the  segment  XaaX  of  the  circumscribed  circle,  and  that  the  centroid 
YBBY  coincides  with,  that  of  the  segment  YhhY  of  the  inscribed 
circle. 

21.  Ci  and  Cj  are  the  centers  of  gravity  of  the  two  portions  of  Fig. 
99.  Show  that  their  distances  from  the  sides  of  the  enclosing  rec- 
tangle, a  X  b,  are  those  marked  in  the  figure.  The  curve  OC  is  a 
parabola.     See  Exercise  5. 

22.  Find  the  centroid  of  the  portion  of  a  right  circular  cylinder 
shown  in  Fig.  100.     C  is  the  centroid.     Its  distance  from  the  axis  of 

the  cylinder  shown  is  — jr — -,  and  from  the  base  is  «  H pj, — 

When  the  oblique  top  cuts  the  base  in  a  diameter  (lower  part  of  Fig. 

100)  the  distance  of  the  centroid  from  the  axis  is  -r^  and  from  the  base 


16 


3ira 
■32' 


18 


274 


CALCULUS 


[§141 


23.  Find  the  centroid  of  the  volume  lying  in  the  first  octant  and 
included  between  the  cylinders  x-  +  y^  =  a^,  x^  +  z*  =  a^ 


4«H 


\ 

C 

• 

1 

1 

" "- 1 

■  > 

a 

__J_ 

^^  ,C 

Fig.  99. 


Fig.  100. 


141.  Theorems  of  Pappus.  Theorem  I.  The  area  oj  the  surface 
generated  by  revolving  an  arc  of  a  plane  curve  about  an  axis  in  its 
plane  and  not  intersecting  it  is  equal  to  the  length  of  the  arc  multi- 
plied by  the  length  of  the  path  described  by  its  centroid. 

Theorem  11.  The  volume  of  the  solid  generated  by  revolving  a. 
plane  surface  about  an  axis  lying  in  its  plane  and  not  intersecting 
its  boundary  is  equal  to  the  area  of  the  surface  multiplied  by  the 
length  of  the  path  described  by  its  centroid. 

Proof  of  I.  Let  ABC,  Fig.  101,  be  an  arc  of  length  L  lying  in  the 
XF-plane.  Then  y,  the  ordinate  of  its  centroid,  is  given  by  the 
equation : 


H 


M 


Whence 


Fig.  101. 


/ 


yds  =  y  L. 


(1) 


(2) 


The  surface  generated  by  revolving  the  arc  ABC  about  the 
X-axis  is  given  by 


§141]  SUCCESSIVE  INTEGRATION  275 


/ 


S  =  2Tr  \  yds.  (3) 

It  follows  then  from  (2)  and  (3)  that 

S  =  2TryL.  (4) 

But  2iry  is  the  length  of  the  circular  path  described  by  the  cen- 
troid  of  the  arc  ABC.     Hence  the  theorem  is  proved. 

Proof  of  II.    Let  ABC,  Fig.  102,  be  a  plane  surface  of  area  A. 
Then  y,  the  ordinate  of  its  centroid,     y 
is  given  by 


.  / 


ydA 

-y  =  ^-j-  (5) 

Whence  ^ 


/ 


ydA=Ay.  (6)  Fig.  102. 


Now  the  volume  of  the  solid  generated  by  the  revolution  of  the 
area  ABC  about  the  X-axis  is 

V  =  2ir  I    j  ydydx  =  2t  I  y  dA.  (7) 

It  follows  from  (6)  and  (7)  that 

V  =  2TrAy.  (8) 

Hence  the  theorem  is  proved.  • 


Exercises 

1.  Find  the  surface  of  the  anchor  ring  generated  by  revolving  the 
circle  x'^  +  {y  —  b)^  =  a^,  a  <  b,  about  the  X-axis. 

2.  Find  the  volume  of  the  anchor  ring  of  Exercise  1. 

3.  Find,  by  using  one  of  the  theorems  of  Pappus,  the  centroid  of  a 
quadrant  of  a  circular  arc,  radius  a. 

Hint.  The  rotation  of  the  arc  about  the  X-axis,  which  coincides 
with  a  radius  drawn  to  one  extremity,  generates  the  surface  S  =  2ira'^. 
Then,  by  (4), 

S  =  27ra2  =  2T2/L  =  2-Ky'^' 


276 


CALCULUS 


[§142 


Hence 


y 


2a 


4.  Find,  by  using  one  of  the  theorems  of  Pappus,  the  centroid  of  a 
quadrant  of  a  circular  area. 

142.  Centroid:  Polar  Coordinates.  The  formulas  are  readily 
obtained  for  finding  the  coordinates  of  the  centroid  of  an  area 
bounded  by  a  curve  whose  equation  is  given  in  polar  coordinates. 
The  area  of  the  element  MPQR,  Fig.  88,  is  p  dp  dd,  and  its  moment 
about  the  F-axis  is  pdpdd  p  cos  6  =  p"^  cos  d  dp  dd.     Hence 


Similarly, 


fCp"^  cose  dp  do 
ff pdpdd 

f  fp^  sine  dp  dd 
ff pdpdd 


X  = 


y  = 


If  it  is  advantageous,  the  integration  with  respect  to  d  can  be 
performed  first. 

Exercises 

Find  the  coordinates  of  the  center  of  gravity  of: 
1.  The  area  of  p  =  a(l  +  cos  6).     The  area  of  the  upper  half  of 
the  same  cardioid. 

2.  The  area  of  one  loop  of  p  = 
a  cos  2d. 

3.  A  circular  sector  of  central  angle 
2  a. 

4.  One  quadrant  of  a  circle.  A 
semicircle.  (Obtain  directly  and  also 
use  the  result  of  Exercise  3.) 

5.  The  area  of  a  portion  of  a  cir- 
cular ring,  Fig.  103,  of  radii  R  and 
r,  and  of  central  angle  2a.  Denote 
by  Cr  the  centroid  of  the  sector  of 

radius  R,  and  by  Cr  that  of  the  sector  of  radius  r,  and  by  C  that 
of  the  given  portion  of  the  ring.  Let  the  abscissas  of  these  points  be 
xr,  Xt,  and  x,  respectively,  and  let  the  corresponding  areas  be  denoted 
by  Ar,  Ar,  and  A. 


Fig.   103. 


§143]  SUCCESSIVE  INTEGRATION  277 

Then 


Hence 


Artr  ■\-  Ax    =  ArXr. 

-  _  "^g^fl  -  ^rXr   ^  2  R^-r^  sin  a 
^~  A  3  R^-r^      a   ' 


Obtain  this  directly  by  integration. 

6.  A  segment  of  a  circle  of  radius  r  cut  off  by  a  chord  of  length  c. 
Use  the  method  of  Exercise  5.  The  distance  of  the  centroid  from  the 
center  is 

(.3    ^  2r^  sin^  a 

12A  ~        M      ' 
where  A  =  area  of  the  segment  =  r^  (2  a  —  sin  2  a:). 

143.  Moment  of  Inertia.  Consider  a  system  of  masses,  mi, 
rriz  ■  ■  . ,  rrin,  moving  with  linear  accelerations,  ji,  j^,  •  •  ■ ,  jn, 
respectively.  The  forces  acting  on  these  masses  are  then  m\j\, 
m^ii,  .  .  .,  rrinjn,  respectively;  and  the  sum  of  the  moments  of 
these  forces  about  an  axis  is  equal  to  l^niijiri  where  ri,  rz,  .  .  . 
r„,  respectively,  are  the  moment  arms  of  these  forces  with  respect 
to  this  axis.  If  now  the  masses  are  rigidly  connected  and  rotate 
about  an  axis,  they  have  a  common  angular  acceleration.  Let  the 
common  angular  acceleration  be  denoted  by  a.  Then  ^i  =  ari, 
ji  =  ar2,  .  .  .,  jn  =  oLTn,  whcro  ri,  r2,  .  .  .,  r„  are  the  dis- 
tances of  the  masses  mi,  m^,  .  .  .,  rUn  from  the  axis  of  rota- 
tion. The  sum  of  the  moments,  T^mijiTi,  becomes  a'Zmiri^. 
This  is  the  moment  necessary  to  produce  the  angular  acceleration 
a.  To  produce  unit  angular  acceleration  a  moment  equal  to 
Sm.ri^  is  necessary.  This  moment,  ^miri"^,  is  called  the  moment 
of  inertia,  and  is  denoted  by  the  symbol  /.     Thus 

I  =  SmiV.  (1) 

The  moment  of  inertia  of  the  system  would  be  unchanged  if  the 
n  masses  of  the  system  were  situated  at  a  distance  k  from  the  axis 
of  rotation  such  that 

'Zniik'^  =  'ZmiTi^, 
or 

^  ~     2mi 


278 


CALCULUS 


[§143 


k  is  called  the  radius  of  gyration.  Its  square  is  the  mean  of  the 
squares  of  the  distances  ri,  ri,  .  .  .  ,  r„  with  respect  to  the  mass. 
The  moment  of  inertia  of  a  system  with  respect  to  an  axis  of 
rotation  plays  the  same  role  in  the  discussion  of  a  motion  of  rota- 
tion as  the  total  mass  in  the  discussion  of  a  motion  of  translation. 
In  the  former  case  the  moment  necessary  to  produce  an  angular 
acceleration  a  is  a^niiri^.  In  the  latter  case  the  force  necessary 
to  produce  a  linear  acceleration  j  is  jSm,-. 

The  kinetic  energy  of  a  rotating  system  can  be  expressed  in 
terms  of  its  moment  of  inertia  and  its  angular  velocity.  If  a 
particle  of  mass  m  is  rotating  with  angular  velocity  co  about  an 
axis  and  at  a  distance  r  from  it,  its  kinetic  energy  is  equal  to 

one-half  the  product  of  its  mass 
by  the  square  of  its  linear  veloc- 
ity, i.e.,  to  jWwV^.  And  if 
there  is  a  system  of  particles  of 
masses  mi,  m2,  •  •  • ,  w„,  at  dis- 
tances ri,  r2,  .  .  .,  r„,  respec- 
tively, from  the  axis,  all  rotating 
with  the  angular  velocity  w,  the 
kinetic  energy  of  the  system  is 
equal  to  ^SmiCoVi^  =  ^oo^Ziriiri^ 

=  Wl- 
If  a  rectangular  plate  of  uniform  thickness  ^  and  composed  of 
material  of  uniform  density,  p,  rotate  about  an  axis  through  one 
corner  and  perpendicular  to  its  plane,  its  moment  of  inertia  can  be 
found  by  a  process  of  double  integration.  Let  the  sides  of  the  rec- 
tangle be  a  and  h  and  take  the  origin  at  one  corner.  Fig.  104. 
The  moment  of  inertia  of  the  rectangle  MNPQ  is  approximately 
the  product  of  its  mass,  p^  Ay  Ax,  and  the  square  of  the  approxi- 
mate distance,  s/x^  -\-  y"^,  of  its  mass  particles  from  the  origin. 
That  is,    the  moment   of   inertia  of   MNPQ   is   approximately 

p^(x2  -I-  y^)Ay  Ax. 
That  of  the  strip  EHIJ  is  approximately 

y  =  b  ^^ 

P^  AySo  DC-^'  +  y')  AyAx=p^Ax  i    (x'  +  y')  dy. 

And  the  moment  of  inertia  of  the  entire  plate  is  obtained  by 


y 

J 

I 

Q 

P 

W 

M 

N 

X 

o 

B 

7  J 

1 

Fig.  104. 


§143]  SUCCESSIVE  INTEGRATION  279 

taking  the  limit  of  the  sum  of  the  moments  of  inertia  of  these 
strips  as  Ax  approaches  zero,  viz., 


«/o    Jo 


I  =  P^£to^^x\   {x^  +  y')dy 


b 

(x2  +  2/2)  dy  dz  =  Wia-"  +  6^), 


where  M  —  p^ab,  the  mass  of  the  plate. 

We  have  obtained  the  moment  of  inertia  of  the  plate  by  inte- 
grating over  its  area  the  product  of  the  mass  of  the  element, 
p^  dy  dx,  by  the  square  of  its  distance,  ^Jx^  +  y"^,  from  the  axis  of 
rotation. 

If  instead  of  a  rectangular  plate  we  consider  a  plate  of  any  shape, 
say  that  of  Fig.  84,  its  moment  of  inertia  is  given  by 

(x^  +  y^)  dy  dx.  (2) 

If  the  density,  p,  and  the  thickness,  ^,  are  variable  the  foregoing 
argument  shows  that  they  must  be  written  under  the  integral  sign. 
For,  the  element  of  integration  is  p^{x^  +  y"^)  dy  dx  and  only  when 
p  and  ^  are  constant  can  they  be  taken  out  from  under  the  integral 
sign.     If  p  and  ^  are  variable  we  have 


Pk  {x^  +  y^)  dy  dx.  (3) 

(2)  and  (3)  can  be  written  in  a  form  easily  remembered,  viz., 

I=Jr2dm,  (4) 

where  dm  is  an  element  of  mass,  and  r  is  its  distance  from  the 
axis. 

Sometimes  in  finding  the  moment  of  inertia  of  a  body  it  is 
advantageous  to  choose  the  element  of  mass  so  that  a  single  inte- 
gral will  suffice.     See  for  example  Illustration  2,  below. 

Illustration  1.  Find  /  of  a  right-angled  triangular  plate  whose 
thickness  is  0.5  inch,  and  whose  legs  are  10  inches  and  4  inches, 
about  an  axis  through  the  vertex  of  the  right  angle  and  perpendicu- 


280 


CALCULUS 


[§144 


lar  to  the  plane  of  the  triangle.     The  density  of  the  material  is 
0.03  pound  per  cubic  inch.     See  Fig.  105. 

I  =  0.03-0.5  II  (a;2  +  y"^)  dij  dx. 


0       t/0 


The  student  will  carry  out  the  integration  and  find  the  radius  of 
gyration. 

Illustration  2.  Find  7  of  a  circular  plate  about  an  axis  through 
its  center  and  perpendicular  to  its  plane. .  The  plate  has  a  radius 
of  10  inches.  It  is  2  inches  thick  and  its  density  is  0.04  pound 
per  cubic  inch. 

Hint.  Here  it  is  convenient  to  divide  the  plate  into  concen- 
tric rings  of  inner  radius  r  and  of  width  dr.     See  Fig.  106.     The 


Fig.  105. 


Fig.  106. 


volume  of  such  a  ring  is  2-2Trr-dr,  and  its  mass  is  0.04*47rr-dr. 
The  distance  of  this  mass  from  the  axis  is  r.     Hence 


I  =  0.16 


■s: 


dr. 


Also  find  the  radius  of  gyration. 

144.  Transfer  of  Axes.  Theorem.  The  moment  of  inertia  of  a 
body  about  any  axis  is  equal  to  its  moment  of  inertia  about  a  parallel 
axis  through  the  centroid,  increased  by  the  product  of  the  jnass  by 
the  square  of  the  distance  between  the  axes. 

Let  AB,  Fig.  107,  be  the  axis  about  which  the  moment  of  inertia 
is  desired.  Choose  a  system  of  rectangular  axes  such  that  the 
origin,  0,  is  at  the  centroid,  such  that  the  Z-axis  is  parallel  to  AB, 


§145] 


SUCCESSIVE  INTEGRATION 


281 


and  such  that  the  ZF-plane  contains  the  line  AB.  Consider  an 
element  of  mass,  dm,  at  P.  The  moment  of  inertia  of  the  body 
about  AB  is  then 

I  =  fiPBy  dm  =  fUy  -  dy  +  x^l  dm, 
or  I  =   j  {x"^  -\-  y"^)  dm  —  2d  j  ij  dm  -{■  d^  j  drn.  (1) 

The  first  term  of  the  right-hand  side  of  (1)  is  the  moment  of 
inertia,  /„,  of  the  body  about  the  Z-axis,  an  axis  through  the  cen- 

troid.     The  second  term,    I  y  dm,  is  the  moment  of  the  body 

with  respect  to  the  XZ-plane,  a  plane  passing  through  its  centroid. 


Fig.  107. 


__       \y  dm 


I   dm 


Since   y  =  0,  j  y  dm   =   0.     The   last   term,   d"^  j  dm,  is  d^M, 
where  M  is  the  mass  of  the  body.     Hence 

I  =  !„  +  Md2. 

145.  Moment  of  Inertia  of  an  Area.  We  have  spoken  of  the 
center  of  gravity  of  an  area  quite  apart  from  any  idea  of  mass  and 
have  stated  that  this  is  a  useful  conception  in  the  study  of  mechan- 
ics. In  the  same  way  the  solution  of  some  problems  in  mechanics 
requires  the  moment  of  inertia  of  an  area  quite  apart  from  any  idea 
of  mass. 


282 


CALCULUS 


[§145 


The  moment  of  inertia  of  a  plane  area  about  an  axis  through  the 
origin  and  perpendicular  to  its  plane  is  defined  by  the  integral. 

I  =  J  J  (x2  +  y2)  dy  dx  =  Jf  (x2  +  y2)  dx  dy. 

The  theorem  on  the  transfer  of  axes  holds  in  this  case  if  the  word 
"area"  is  substituted  for  the  word  "mass." 


Exercises 

Find  the  moment  of  inertia  of  the  following : 

1.  A  rectangle  of  sides  a  and  b  about  one  comer.  (See  Fig.  104.) 
About  the  centroid.  About  one  base.  About  a  line  parallel  to  one 
base  and  passing  through  the  centroid. 

2.  A  right  triangle,  legs  a  and  b,  about 
one  of  the  legs.     About  a  line  through  the 

^^  centroid  parallel  to  this  leg. 

3.  The   area   of  a   circle  about   an  axis 
'  through  a  point  on  its  circumference  and 

perpendicular  to  its  plane.     See  Illustration 
2,  §143. 

4.  The  area  of  a  circle  about  a  diameter. 
Hint. 


I  = 


X? 


2xdy. 


Fig.  108. 


5.  The  area  of  a  circle  about  a  tangent 
line. 

6.  The  area  between  y  =  x^  and  y^  =  x 
about  an  axis  through  the  origin  perpen- 
dicular to  the  XF-plane. 

7.  A  uniform  bar  of  length  L  and  linear 
density   p   about  an   axis  through  one  end 

perpendicular  to  the  bar.     Find  /  about  a  parallel  axis  through  the 
middle  point  of  the  bar. 

8.  A  bar  of  length  L,  whose  density  is  proportional  to  the  distance 
from  one  end,  about  an  axis  perpendicular  to  the  bar  through  the 
end  of  least  density. 

9.  A  slender  uniform  rod.  Fig.  108,  about  a  line  through  its  middle 
point  and  making  an  angle  a  with  the  rod. 

Ans.  I  —  ^s'lnL^  sin''  a,  where  m  is  the  mass  and  L  is  the  length 
of  the  rod. 


§146]  SUCCESSIVE  INTEGRATION  283 

Hint.     Denote  by  p  the  linear  density.     Then 

I  =  P  i  x^  sin^a  dx,  with  proper  limits. 

10.  The  rod  of  Exercise  9  about  a  parallel  axis  through  one  end. 

11.  A  wire  bent  into  the  form  of  a  circular  arc,  Fig.  109,  about  the 
origin.  Also  find  the  moments  of  inertia,  Ix  and  ly  about  the  X-  and 
F-axes,  respectively. 


/- 
/- 


7  =    I     r^rdB; 
Ix  =    I     r^sin^erdd', 


/„  =    I     r^  cos^e  r  de. 


Fig.  109. 


12.  A  triangle  of  base  h  and  altitude  h  about  an  axis  through  the 
vertex  parallel  to  the  base.  Divide  the  area  into  strips  parallel  to  the 
base  and  of  width  dx.  The  axis  of  x  is  drawn  from  the  vertex  perpen- 
dicular to  the  base. 

_    ,    x^hx  dx 


f 


13.  A  triangle  about  a  line  through  the  center  of  gravity  parallel 
to  the  base.     Use  the  result  of  Exercise  12. 

14.  The  area  of  the  ellipse  ^  +  rj    =1  about  the  major  axis.     (Use 

single  integration.)     About  the  minor  axis.     About  the  origin. 

146.  Moment  of  Inertia:  Polar  Coordinates.  The  moment  of 
inertia  of  the  element  r  dr  dd  about  an  axis  through  the  origin  is 
r^r  dr  dd.     Hence  the  moment  of  inertia  of  an  area  is 

/  =  J Jr3  dr  dd, 

with  proper  limits.     If  the  moment  of  inertia  of  a  plate  is  required, 
r'  dr  dd  is  to  be  multiplied  by  p,  the  density  per  unit  area. 

Exercises 

Find  the  moment  of  inertia  of  the  following : 

1.  The  area  of  the  cardioid  p  =  a  (1  +  cos  6)  about  an  axis  through 
the  origin  perpendicular  to  the  plane  of  the  cardioid.  About  the 
initial  line. 


284  CALCULUS  [§147 

2.  The  area  of  one  loop  of  p  =  o  cos  2d  about  the  initial  line. 

3.  A  circular  sector  of  central  angle  2a  about  the  radius  of 
symmetry. 

4.  The  arc  of  the  sector  of  Exercise  3  about  the  radius  of  sym- 
metry. 

5.  The  area  of  Exercise  3,  about  an  axis  through  the  center  of  the 
circle  and  perpendicular  to  the  plane  of  the  sector.  About  a  parallel 
axis  through  the  centroid. 

147.  Moment  of  Inertia  of  a  Solid. — We  wish  to  find  the  moment 
of  inertia  of  the  solid  of  Fig.  86,  about  the  Z-axis.  The  moment  of 
inertia  of  the  element  of  mass,  p  dz  dy  dx,  about  the  Z-axis  is  equal 
to  pix"^  +  7/^)  dz  dy  dx.  The  total  moment  of  inertia  of  the  solid 
about  this  axis  is  the  integral  of  this  element  throughout  the  solid. 
Hence, 

/.  =    I      I       I  p(x2  +  7/2)  dz  dy  dx.  (1) 


Similarly 


r,  -  f  r  r 

J  a    Jy^    Jz  = 

n"2    ni  =  fix. 
J    Jz  =  0 

Jo,    Jy,    Jz  = 


y) 
/«=    I      j       I  p{y^  +  z^)  dz  dy  dx,  (2) 


•z  -  fix,  y) 

p{z^  +  x2)  dz  dy  dx.  (3) 

0 

If  the  solid  be  regarded  as  a  geometrical  volume  of  density  1, 
the  p's  disappear,  and  the  formulas  (1),  (2),  and  (3)  can  be  written 

/.  =  Jffi^'  +  y')  dz  dy  dx,  (4) 

I.  =  Jffiy'  +  2')  dz  dy  dx,  (5) 

ly  =  ///(2'  +  x^)  dz  dy  dx.  (6) 

Let 

/„,  =    III  px^  dz  dy  dx,  (7) 

Lx  =  JJJpy'^  dz  dy  dx,  (8) 

hv  =    fffp^^  dz  dy  dx.  (9) 

The  quantities  (7),  (8),  and  (9)  will  be  called  the  moments  of 
inertia  of  the  solid  with  respect  to  the  FZ-plane,  the  XZ-plane, 
and  the  XF-plane,  respectively.    They  are  the  integrals  of  the 


§147]  SUCCESSIVE  INTEGRATION  285 

product  of  the  element  of  mass  by  the  square  of  its  distance  from 
the  respective  planes.  They  can  very  frequently  be  found  by  a 
single  integration  by  taking  as  element  a  plane  lamina  between 
two  planes  parallel  to  the  plane  with  respect  to  which  the  moment 
is  computed.  If  this  is  the  case  the  moment  of  inertia  about  the 
coordinate  axes  can  easily  be  found  by  noting  that  from  the  equa- 
tions(l),  (2),  (3);  and  (7),  (8),  (9): 


Iz 

= 

Iy^ 

+  Ixz, 

I. 

= 

Ixz 

+  Ixy, 

ly 

= 

Ixy 

+  Iyz. 

That  is,  the  moment  of  inertia  about  the  Z-axis  is  equal  to  the  sum 
of  the  mom£nts  of  inertia  with  respect  to  the  YZ-  and  XZ-planes,  and 
so  on. 

In  general,  the  moment  of  inertia  of  a  body  about  an  axis  is  equxil 
to  the  sum  of  its  moments  of  inertia  with  respect  to  two  perpendicular 
planes  which  intersect  in  that  axis. 

In  the  same  way  it  follows  from  the  formula  for  the  moment  of 
inertia  of  an  area,  I  =    I  I  (a;^  +  y^)  dy  dx,  that  the  sum  of  the 

moments  of  inertia  of  an  area  {or  a  plate)  about  two  perpendicular 
axes  is  equal  to  its  moment  of  inertia  about  an  axis  perpendicular 
to  the  plane  of  these  axes  through  their  point  of  intersection. 

Illustration  1.     Find  the  moment  of  inertia  of  the  ellipsoid 


X  V  z 

—i  +  T^-{ — s  =  1  about  each  of  its  axes. 
a^       0^       c'- 

First  Method. 


y 


/.  =   8j       j        ^  a^-^'\^       a^       %J^  +  Z^)  dz  dy  dx. 

Carry  out  the  integration  far  enough  to  see  that  it  is  not  simple 
and  then  note  the  relative  simplicity  of  the 

Second  Method.  Compute  Ixz,  the  moment  of  inertia  with 
respect  to  the  XZ-plane.  Take  as  element  of  integration  the  ellip- 
tical plate  cut  out  by  the  planes  y  =  y  and  y  =  y  -\-  Ay.  The 
equation  of  the  intersection  of  the  ellipsoid  and  the  plane  y  =  y  is 
x^ 2^  _ 


.»(i-fi)   .<!-«;) 


286  CALCULUS  [§147 

Now  the  area  of  an  ellipse  is  t  times  the  product  of  its  semi- 
major  and  semi-minor  axes.     Hence  the  area  of  this  ellipse  is 

The  volume  of  the  elliptical  plate  in  question  is 

irac  [l  -  p-j  dy 

and  its  moment  of  inertia  with  respect  to  the  XZ-plane  is 

xacy^  (l  -  p-j dy. 

The  total  moment  of  inertia  of  the  ellipsoid  with  respect  to  this 
plane  is  then 

I,,  =  Tracfy^[l-pjdy 
=  27roc  ( 


/63       5  A 


3        5/  16 

I XV  can  be  written  down  at  once  as 

-.     _  47ra6c' 

Then 

T  TIT  ^Trahc  ,,  „    . 

We  can  write  down  at  once  by  interchanging  letters: 

Illustration  2.  Find  the  moment  of  inertia  of  a  right  circular 
cone  about  a  line  through  its  vertex  perpendicular  to  its  axis,  if 
the  radius  of  the  base  is  b  and  the  altitude  is  h.  Choose  the  vertex 
as  origin  and  the  axis  of  the  cone  as  axis  of  x.     Consider  the  plate 

bx 
of  radius  r  cut  out  by  the  planes  x  =  x  and  x  =  x  +  dx.     Its 


§147]  SUCCESSIVE  INTEGRATION  287 

moment  of  inertia  about  a  diameter  parallel  to  the  axis  of  rotation 

through  the  vertex  is  equal  to  — j-rz — .     (See  expression  for  /  of 

a  circular  plate  about  a  diameter,  Exercise  4,  §145).  Then  the 
moment  of  inertia  of  this  plate  about  the  axis  of  rotation  through 
the  vertex  is  equal  to  this  moment  of  inertia  about  an  axis  through 
the  centroid  increased  by  its  volume  (mass  if  density  =  1)  multi- 
plied by  the  square  of  the  distance  between  this  axis  and  the  par- 
allel axis  through  the  vertex  (see  §144)  i.e.,  to 

Tb*x*dx        irb^x'^dx  ^ 
W      "^       P      ^  ■ 

And  the  total  moment  of  inertia  of  the  cone  about  the  axis  through 
the  vertex  is  equal  to  the  integral  of  this  moment  of  inertia  from 
X  =  0  to  X  =  h.    That  is 


SX 


■jrb^h    ,   Trb%^        Trb-h ,, „    ,    ^,,. 
"20-  +-5-  = -20"  (^^  +  ^'^^)- 


Exercises 


Find  the  moment  of  inertia  of: 

1.  The  cone  of  Illustration  2,  about  a  parallel  axis  through  the 
centroid  of  the  cone.     About  a  diameter  of  the  base. 

2.  A  right  circular  cylinder,  the  radius  of  whose  base  is  r,  and  whose 
altitude  is  h,  about  a  diameter  of  one  base.  About  a  parallel  axis 
through  the  centroid. 

3.  A  rectangular  parallelopiped  with  edges  a,  b,  and  c,  about  an 
axis  through  the  centroid  parallel  to  one  edge. 

4.  A  right  circular  cylinder  about  its  axis. 

5.  A  hollow  right  circular  cylinder  of  outer  radius  R,  inner  radius  r, 
and  altitude  h,  about  its  central  axis.  About  a  diameter  of  one  base. 
About  a  diameter  of  the  plane  section  through  the  centroid  perpen- 
dicular to  the  axis. 

6.  A  right  rectangular  pyramid  of  base  a  X  b  and  of  altitude  h, 
about  an  axis  through  the  centroid  parallel  to  the  edge  a.  About  an 
axis  through  the  vertex  and  the  center  of  gravity. 

Ans.     /.  =  ^  (b-  +  Ih^).     h=^  {a'  +  &=). 


288  CALCULUS  [§147 

7.  A  frustum  of  a  right  cone  about  its  axis  if  the  radius  of  the  large 
base  is  R,  that  of  the  small  base  is  r,  and  the  altitude  is  h. 

R^  —  r^ 

Ans.     I  =  -fifTrh  -fz 

R  —  r 

8.  A  hollow  sphere  about  a  diameter,  if  the  outer  radius  is  R  and 
the  inner  radius  is  r. 

9.  A  paraboloid  of  revolution,  the  radius  of  whose  base  is  r  and 
whose  height  is  h,  about  the  axis  of  revolution. 

Ans.     I  =  lirr*h. 

10.  The  anchor  ring  generated  by  revolving  the  circle 
[x  -  R]^  +  y^  =  r^  about  the  F-axis. 

Ans.     h  =  TT^RrKR^  +  fr").     ly  =  27rmr\R^  +  fr^). 

11.  A  right  circular  cone  about  its  axis. 

12.  A  right  elliptical  cylinder  of  height  L,  and  having  the  semi-major 
and  semi-minor  axes  of  its  cross  section  equal  to  a  and  b,  respectively, 
about  an  axis  through  the  centroid  parallel  to  b. 

13.  A  quadrant  of  a  circular  plate  about  one  of  its  bounding  radii. 

14.  An  equilateral  triangle  of  side  2a,  about  a  median.  About  a 
line  through  a  vertex  perpendicular  to  one  of  the  sides  through  that 
vertex. 


CHAPTER  XVI 


CURVATURE.    EVOLUTES.    ENVELOPES 


y 

1 

qI 

A- 

Pf        /r+t^T 

o 

y 

/ 

Fig.  110. 


148.  Curvature.  Let  PT  and  QT,  Fig.  110,  be  tangents  drawn 
to  the  curve  APQ  at  the  points  P  and  Q,  respectively.  Denote  the 
length  of  the  arc  PQ  by  As  and 
the  angles  of  inclination  of  PT 
and  QT'  to  the  positive  direc- 
tion of  the  X-axis  by  t  and  t 
-j-  Ar,  respectively.  At  gives  a 
rough  measure  of  the  deviation 
from  a  straight  line,  of  that  por- 
tion of  the  arc  of  the  curve  be- 
tween the  points  P  and  Q.  The 
sharper  the  bending  of  the  curve 
between  the  points  P  and  Q  the 

greater  is  At  for  equal  values  of  As.      The  average  curvature  of 
the  curve  between  the  points  P  and  Q  is  defined  by  the  equation 

At 
Average  Curvature  =  't~''  (1) 

The  average  curvature  of  a  curve  between  two  points  P  and  Q 
is  the  average  change  between  these  points,  per  unit  length  of 
arc,  of  the  inclination  to  the  X-axis  of  the  tangent  line  to  the 
curve.  Or,  more  briefly,  the  average  curvature  is  the  average 
change  per  unit  length  of  arc,  in  the  inclination  of  the  tangent  line. 

The  curvature  at  P  is  defined  as  the  limit  of  the  average  curvature 
between  the  points  Q  and  P  as  Q  approaches  P.  On  denoting  the 
curvature  by  K,  we  have  in  accordance  with  the  definition, 

^'       ^'  (2) 


K  =    1™ 

■^  As  =  0  As 


ds 


The  curvature  at  a  point  P  is  then  the  rate  of  change  at  this  point  of 
the  inclination  of  the  tangent  line  per  unit  length  of  arc.      The 
19  289 


290 


CALCITLUS 


I  §151 


curvature  is  a  measure  of  the  amount  of  bending  of  a  curve  in 
the  vicinity  of  a  point, 

149.  Curvature  of  a  Circle.     It  is  clear  that  the  average  curva- 
ture of  a  circle,  Fig.  Ill,  is 


At  _   At 
As       r  At 


Hence  the  average  curvature  is  independent  of  As  and  conse- 
quently the  curvature,  the  limit  of  the  average  curvature  as 
As  approaches  zero,  is 


A'  = 


(1) 


Tfie  curvature  of  a  circle  is  constant  and  equal  to  the  reciprocal  of  its 

radius. 

150.  Circle  of  Curvature.    Radius  of  Curvature.     Center  of 

Curvature.  Through  any  point  P  of  a  curve  infinitely  many  cir- 
cles can  be  drawn  which  have  a 
common  tangent  wdth  the  curve  at 
P  and  whose  centers  are  on  the  con- 
cave side  of  the  curve.  Of  these 
circles  there  is  one  whose  curvature 
is  equal  to  that  of  the  curve  at  P, 
i.e.,  one  whose  radius  is  equal  to  the 
reciprocal  of  the  curvature  at  P. 
This  circle  is  called  the  circle  of 
curvature  at  the  point  P.  The 
radius  of  this  circle  is  called  the 

radius  of  curvature,  and  its  center  the  center  of  curvature,  of  the 

curve  at  the  point  P.     The  radius  of  curvature  is  denoted  by  R 

and,  in  accordance  with  (2),  §148,  its  length  is 


Fig.  111. 


R  = 


K 


ds 
dr* 


(1) 


151.  Formulas  for  Curvature  and  Radius  of  Curvature:  Rec- 
tangular Coordinates.  For  obtaining  the  curvature  at  any  point 
on  the  curve  y  =  /(x),  we  shall  now  develop  a  formula  involving 


§151]  CURVATURE.     EVOLUTES.     ENVELOPES  291 

the  first  and  second  derivatives  of  y  with  respect  to  x.    The  above 
formula  for  curvature  K  can  be  written 


(1) 


dr                dr 

dx                dx 

dx       \1  +  [dx) 

Since 

.       ,dy 

T  =  tan-^  -j-y 

dx 

d'y 
dr             dx"^ 

Consequently 

and,  by  (1),  §150, 

[^M^Q? 

,  [-(S)r 

^  -              d2y 
dx2 

(2) 


(3) 


We  shall  understand  by  K  and  R  the  numerical  values  of  the 
right-hand  members  of  (2)  and  (3),  respectively,  since  we  shall 
not  be  concerned  with  the  algebraic  signs  of  K  and  R. 

Illustration.     Find  the  curvature  oi  y  =  x^. 

^  =  2x     ^  =  2 
dx  '    dx^ 

Substitution  in  formula  (2)  gives 

2 


K  = 


(1  +  4x2)^ 


From  this  expression  it  is  seen  that  the  maximum  curvature 
occurs  when  x  is  zero,  and  that  the  curvature  decreases  as  x 


292  CALCULUS  [§152 

increases    in    numerical    value.     When   a;  =  0,    K  =  2.     When 


>   —    X    -1-,    -IV    — 

25 

Exercises 

Find  the  curvature  and  radius  of  curvature  of  each  of  the  curves: 

l.y  =  2x-  xK 

A.  y  =  x^  —  x^. 

7.  2/  -  3x*. 

2.  2/  =  xi 

5.  y  =  -.• 

8.  y  =  x~^. 

-y-l- 

6.  2/  =  Vx. 

Vx 

10.  If  p  =  /(&)  is  the  equation  of  a  curve  in  polar  coordinates,  show 
that 

„      ^    +Hd"gJ     -^d¥^ 
K  —  ,      • 


Hint. 


['-(^^)T 


_  dr  _  d^ 

"~  ds  ~  ds 

d« 

T  =  e  +ip.  (See  Fig.  72.) 

til-  _  d^ 

50  ~      "^  do" 


Obtain  -jr  from  the  relation 
dff 


d0 


IS  given  in 


de 

152.  Curvature:  Parametric  Equations.  If  the  equation  of  a 
curve  is  expressed  in  parametric  form,  z  =  f{t),  y  =  F{i),  the  cur- 
vature can  be  found  by  differentiating  x  and  y  and  substituting 
in  (2),  §151.     t  can  be  eliminated  from  the  result  if  desired. 

Illustration  1 .    li  x  =  t  and  y  =  t^, 

^  ^  2«  and  ^  =  ^^  —  =  2 
dx  '         dx'         dt    dx 


§153]  CURVATURE,    EVOLUTES.    ENVELOPES  293 

Hence 

2  2 


Illustration  2.     Find  the  curvature  of  the  ellipse  x  =  a  cos  t, 
y  =  6  sin  t. 


dy  b 

-J-  = cot  t. 

ax  a 


d-hj  b  d      .dt        &       „, 

3-:  = -V,  cot  t  -3—  =  —  csc^  t 

dx^  a  at  ax        a 


r— T      = iCSC^^ 

a  sin  t_\  a^ 


-  -cscH 
„  a^  —  ab  —  ab 

h.  = 


[l+^'cot^f]^      (a'sinH  +  b^cosH)^       g  ^2  +  ^J  ^^1 


a*b* 


ib*x^  +  a*y^Y 

Exercises 

1.  Find  the  curvature  of  the  curve  x  =  a  cosh  I,  y  —  a  sinh  t. 

2.  Find  the  curvature  of  the  curve  x  =  a{t  —  sin  0) 
y  =  a{\  —  cos  0- 

153.  Approximate   Formula   for   the    Curvature.     If   a   curve 

dy 
deviates  but  little  from  a  horizontal  straight  line,  -r-  is  small  and 

(dv\  2 
^1    in    formula  (2),    §151,  is  very  small   compared   with   1. 

Hence  the  denominator  differs  very  little  from  1  and  the  formula 
becomes  approximately 

^=S-  (') 

This  approximate  formula  for  K  is  frequently  used  in  mechanics 
in  the  study  of  the  flexure  of  beams.     The  slope  of  the  elastic 

d'^y 
curve  of  a  beam  is  so  small  that  -v-^  can  be  used  for  the  curva- 
ture without  appreciable  error. 

The  approximate  formula  for  the  radius  of  curvature  R  is 

7?-  ^ 

^~  d^  (2) 

dx^ 


294 


CALCULUS 


[§154 


154.  Center  of  Curvature.  Evolute.  Formulas  will  now  be 
obtained  for  the  coordinates  of  the  center  of  curvature  of  a  curve 
corresponding  to  any  point  P.  Let  the  coordinates  of  P  be  a;  and 
y.  Denote  by  a  and  j8  the  coordinates  of  the  center  of  curvature 
of  the  curve  at  this  point.  There  are  four  cases  to  be  considered. 
See  Fig.  112,  a,  b,  c,  d. 


Fig.  112. 


In  Fig.  112,  a, 


a  =  OM  =^  ON  -  HP  =  X  -  Rsinr, 
/3  =  MC  =  NP  +  HC  =  y  +  R  cost. 


Since 


tanr  = 


dy 
dx 


cost  = 


NRS-f 


;    sinT 


dy 

dx 


/dy\^ 
\dx) 


§154]  CURVATURE.     EVOLUTES.     ENVELOPES  295 

Consequently, 


^-m 


«=--f-d^f^'       (1) 


and 


1  + 


(^) 


^  =  y  +  — d^-  (2) 

The  student  can  show  that,  since  -7—  is  negative  for  a  descending 

curve  and  positive  for  an  ascending  curve,  and  since  -r^  is  positive 

when  a  curve  is  concave  upward  and  negative  when  a  curve  is 
concave  downward,  formulas  (1)  and  (2)  hold  for  the  three  curves 
represented  in  Fig.  112,  b,  c,  d. 

Illustration.  Find  the  coordinates  of  the  center  of  curvature 
corresponding  to  any  point  on  the  curve  y  =  ±  2  y/~x.  Only  the 
positive  sign  will  be  used.  If  the  negative  sign  is  used  it  will 
only  be  necessary  to  change  the  sign  of  ^. 

dy  ^     1 

dx  y/^ 

^    =      _    J_ 

dx^  2a;i* 

1 


1 


1  + 


2x1 


=  y 1 —  =  y 


-  2  V^(x  +  1)  =  2/  -  2/  [I  +  1]  =  - 


2x3 

The  equation  of  the  locus  of  the  center  of  curvature  is  obtained 

by  eliminating  x  and  y  from  the  equations  for  a  and  /3  and  the 

equation  of  the  original  curve.     Thus 

0=  ~  2  /.ox' 

X  =  — y— ;    y  =  -  (4)8)'  . 


296  CALCULUS  [§155 

Substituting  in  y"^  =  4a;,  we  obtain 

^^  =  A(«  -  2)», 
the  equation  of  the  locus  of  the  center  of  curvature.     This  is  the 
equation  of  a  semi-cubical  parabola  whose  vertex  is  at  the  point 
(2,0). 

The  locus  of  the  center  of  curvature  corresponding  to  points  on  a 
curve  is  called  the  evolute  of  that  curve.  Its  equation  is  easily 
obtained  in  many  cases  by  eliminating  x  and  y  from  equations  (1) 
and  (2)  and  the  equation  of  the  original  curve.  Otherwise  (1) 
and  (2)  constitute  its  parametric  equations,  a  and  /3  being  ex- 
pressed in  terms  of  the  parameters  x  and  y. 

Exercises 

1.  Find  the  evolute  of  t/  =  4x^. 

2.  Find  the  evolute  of  the  ellipse 


I* 


^  ^yi  -  1 

a^  "^  6*       ^' 

Hint      It  will  be  found  that 

(a*  -  &2)x'                    (a«  -  6*)y« 
«  =  ^, ;     ^  = ^, 

Whence 

Elimination  gives 

iaa)i  +  (6^)5  =  (o2  -  62)5. 

3.  Find  the  parametric  equations  of  the  evolute  of  the  cycloid, 

X  =  a{9  —  sin  9), 
y  =  a(l  —  cos  e). 
Ans.  a  =  a(0  +  sin  6),  0  =  —  a(l  —  cos  d).     Show  that  the  evo- 
lute is  an  equal  cycloid  with  its  cusp  at  the  point  ( —  wa,  —  2a). 

4.  Find  the  equation  of  the  evolute  of 

X  =  a(cos  9  +  9sm  e), 
y  =  o(sin  0  —  ^cos^). 
Ans.  a  =  a  cos  9,     /3  =  a  sin  9.     Discuss. 

155.  Envelopes.  If  the  equation  of  a  curve  contains  a  constant 
c,  infinitely  many  curves  can  be  obtained  by  assigning  different 
values  to  c.     Thus 

(x  -  c)2  +  2/2  =  a2  (1) 


§155] 


CURVATURE.     EVOLUTES.     ENVELOPES 


297 


is  the  equation  of  a  circle  of  radius  a  whose  center  is  at  (c,  0).  By- 
assigning  different  values  to  c  we  get  a  system  of  equal  circles 
whose  centers  lie  on  the  X-axis.  A  constant  such  as  c,  to  which 
infinitely  many  values  are  assigned,  is  called  a  parameter.  A 
constant  such  as  a,  which  is  thought  of  as  taking  on  only  one  value 
during  the  whole  discussion,  is  called  an  absolute  constant.  We  say 
that  equation  (1)  represents  Si  family  of  circles  or  a  system  of  circles 
corresponding  to  the  parameter  c. 

The  general  equation  of  a  family  of  curves  depending  upon  a 
single  parameter  can  be  written  in  the  form, 


fix,  y,  c)  =  0. 


(2) 


Exercises 
State  the  family  of  curves  represented  by  the  following  equations 


contammg  a  parameter: 

1.  y  =  x^  +  c, 

2.  y  =  mx  +  b, 

3.  y  =  mx  +  b, 

*•  a»  ^  &»       ^' 
6.  2/*  =  m(x  +  m), 
6.  x^  +  y^  =  a\ 


c  being 
b  being 
m  being 

a  being 

TO  being 
a  being 


the  parameter, 
the  parameter, 
the  parameter. 

the  parameter. 

the  parameter, 
the  parameter. 


Consider  again  the  family  of  circles  (1).  Two  circles  of  the 
family  corresponding  to  the  values,  c  and  c  -f-  Ac,  of  the  parameter 
intersect  in  the  points  Q  and  Q', 
Fig.  113.  We  seek  the  limiting 
positions  of  these  points  of  in- 
tersection as  Ac  approaches  zero. 
Clearly  they  are  the  points  P 
and  P',  respectively,  on  the  lines 
y  =  +a.  Such  a  limiting  posi- 
tion of  the  point  of  intersection 
of  two  circles  of  the  family  is 
called  the  point  of  intersection  of 
two  "consecutive"  circles  of  the 

family.     In  general,  the  limiting  position!  f  the  point  of  intersec- 
tion of  two  curves,  f{x,  y,  c),  f  (x,  y,c-\-  Ac),  of  a  family,  as  Ac 


298  CALCULUS  [§155 

approaches  zero,  is  called  the  point  of  intersection  of  "consecu- 
tive" curves  of  the  family. 

In  the  case  of  the  family  of  circles  (1)  the  locus  of  the  points  of 
intersection  of  "consecutive"  circles  is  the  pair  of  straight  lines 
y  =  +a.  This  locus  is  called  the  envelope  of  the  family  of  circles. 
In  general,  the  envelope  of  a  family  of  curves  depending  upon  one 
parameter  is  the  locus  of  the  points  of  intersection  of  "consecutive" 
curves  of  the  family.  It  will  be  shown  in  a  later  chapter  that  the 
envelope  of  a  family  of  curves  is  tangent  to  every  curve  of  the 
family. 

Exercise 

Draw  a  number  of  lines  of  the  family 

X  cos  6  +  y  sin  d  =  p, 

where  9  is  the  parameter,  and  sketch  the  envelope. 

A  general  method  of  obtaining  the  envelope  of  a  family  of 
curves  will  now  be  given. 

The  equation  of  a  curve  of  the  family  is 

f(x,  y,  c)  =  0,  (3) 

where  c  has  any  fixed  value.  The  envelope  is  the  locus  of  the 
limiting  position  of  the  point  of  intersection  of  any  curve  (3)  of 
the  family  with  a  neighboring  curve,  such  as 

fix,  y,  c  +  Ac)  =  0,  (4) 

as  the  second  curve  is  made  to  approach  the  first  by  letting  Ac 
approach  zero.  The  coordinates  of  the  points  of  intersection  of 
the  curves  representing  equations  (3)  and  (4)  satisfy 

fix,  y,  c  +  Ac)  -  fix,  y,  c)  =  0.  (5) 

Then  they  satisfy 

fix,y,c  +  Ac)  -fix,y,c) 


Ac 


=  0,  (6) 


since  Ac  does  not  depend  on  either  x  or  y.     Then  the  coordinates 
of  the  limiting  positions  of  these  points  of  intersection  satisfy 

lim  fix,y,c  +  Ac)  -fix,y,c)  _ 
Ac  A  0  ^c  ~ 


§155]  CURVATURE.     EVOLUTES.     ENVELOPES  299 

The  first  member  of  this  equation  is  the  derivative  of  /(x,  y,  c) 
with  respect  to  c.     It  may  be  written  in  the  form, 

df(x,  y,  c)  _  , 

dc        ~  ^-  ^'' 

The  differentiation  is  performed  with  respect  to  c,  x  and  y  being 
treated  as  constants.  The  point  of  intersection  also  lies  on  (3). 
Hence  the  equation  of  its  locus  is  obtained  by  eliminating  c 
between  (3)  and  (7). 

Illustration  1.  Find  the  equation  of  the  envelope  of  the 
family  of  circles,  {x  —  cy  -{-  y^  =  a^,  c  being  the  parameter. 

The  equation  of  the  curve  written  in  the  form  f{x,  y,  c)  =  0  is 

{x  -  cY  +  2/2  _  a2  =  0.  (I) 

Differentiating  with  respect  to  c, 

-  2{x  -  c)  =  0.  (II) 

The  elimination  of  c  between  (I)  and  (II)  gives 

or 

y  =  ±a, 
as  the  envelope. 

Illustration  2.    Find    the  equation  of    the  envelope  of    the 
family  of  lines,  x  cos  6  -^  y  sin  d  =  p,  6  being  the  parameter. 
On  differentiating  the  first  member  of 

X  cos  d  -\-  y  sin  d  '-  p  =  0  (I) 

with  respect  to  d  we  obtain 

—  X  sin  6  -\-  y  cos  d  =  Q.  (II) 

The  result  of  eliminating  d  between  (I)  and  (II)  is 

^2   _j_  ^2    _    p2^ 

a  circle  of  radius  p  about  the  origin  as  center. 

Exercises 

P 
1.     Find  the  envelope  of  the  family  of  straight  lines  y  —  mx  -\ — ' 

m 

where  m  is  the  parameter.     Draw  figure. 


300  CALCULUS  [§156 


2.  Find  the  envelope  of  the  family  of  lines  y  =  mx  +  o\/l  +  m', 
where  m  is  the  parameter.     Draw  figure. 

3.  Find  the  envelope  of  the  family  of  parabolas  y^  =  c(x  —  c),  c 
being  the  parameter. 

4.  Find  the  envelope  of  the  family  of  lines  of  constant  length  whose 
extremities  lie  in  two  perpendicular  lines. 

6.  Find  the  envelope  of  y  =  px  —  -p^,  p  being  the  parameter. 
Draw  figure. 

6.  Find  the  envelope  of  the  family  of  curves  (x  —  c)^  +  2/^  =  4pc, 
c  being  the  parameter.     Draw  figure. 

7.  The  equation  of  the  path  of  a  projectile  fired  with  an  initial 
velocity  Vo  which  makes  an  angle  a  with  the  horizontal,  is 

y  —  X  tan  a  —  n — ; 7. — 

"  2v^  cos^a 

Find  the  envelope  of  the  family  of  paths  obtained  by  considering  a  a 

parameter. 

Wo*      gx^ 
Ans.  V  ==  -ji—  —  77—! 

8.  The  equation  of  the  normal  to  y^  =  4x  at  the  point  P,  whose 
coordinates  are  Xi  and  yi,  is 

yi  ,  ■. 

y  -  yi  =  -  2^  (x— xi). 

Since  yi^  =  4xi,  this  may  be  written 

2/iX  +2y  -'^       2yi=  0. 

Find  the  equation  of  the  envelope  oi  the  normals  as  P  moves  along 
the  curve. 

Hint.  On  differentiating  with  respect  to  the  parameter  2/1  we 
obtain 

a.  o  V^^^Z? 

On  substituting  this  value  of  yi  in  the  equation  of  the  normal  and 
squaring  we  obtain 

2       4(x  -  2)^ 

y  =  —27 — 

This  is  the  evolute  of  the  parabola  as  we  have  seen  in  §164. 

156.  The  Evolute  as  the  Envelope  of  the  Normals.  In  Exercise 
8  of  §155  it  was  seen  that  the  evolute  of  a  parabola  is  the  envelope 
of  its  normals.  This  is  true  for  any  curve.  The  result  is  fairly 
evident  from  the  examination  of  the  curves  of  the  exercises  of 


§156]  CURVATURE.     EVOLUTES.     ENVELOPES  301 

§165  and  their  evolutes.     It  will  be  shown  that  the  normals  to  a 
curve  are  tangent  to  its  evolute. 

The  parametric  equations  of  the  evolute  are 

a  =  X  —  Rsinr,  (1) 

^  =  y  +  RcosT.  (2) 

On  differentiating  with  respect  to  the  variable  s,  which  is  per- 
missible since  x,  y,  R,  and  r  are  aU  functions  of  s,  we  obtain 

da       dx       dR   .  _,  dr 

3—  =  3 5—  sm  T  —  /t  cos  r  3-  • 

ds        ds        ds  ds 


d^       dy    ,   dR  „   .       dr 

T"  =  3 r  T'  COST  —  R  smr  -j-- 

ds       ds        ds  ds 


Now 


=    COST, 


dx 
ds 
dy 

ds  ~  R' 

Then  the  foregoing  equations  become 
da  dR   . 

ds  ds  ' 

d^      dR 

3—  =   3—  COS  T. 

ds       ds 
Hence  the  slope  of  the  tangent  to  the  evolute  is 

^  =  -cotT  (3) 

da 

Therefore  the  tangent  to  the  evolute  is  parallel  to  the  normal  to 
the  curve  at  the  point  (x,  y)  to  which  (a,  jS)  corresponds.  But  the 
normal  to  the  curve  at  (x,  y)  passes  through  (a,  /3).  Hence  it  is 
tangent  to  the  evolute  at  this  point. 

It  can  also  be  shown  that  if  Ci  and  C2,  Fig.  114,  are  the  centers 
of  curvature  corresponding  to  the  points  Pi  and  P2,  the  length  of 
the  arc  C1C2  of  the  evolute  is  equal  to  the  difference  in  the  lengths 
of  the  radii  of  curvature,  Ri  and  R^.  For,  from  the  above  values 
of  da  and  d^  it  follows  that 

Vda^  +  diS*  =  dR. 


302  CALCULUS  [§157 


But  VdoM-d^  is  the  differential  of  the  arc  of  the  evolute.  Call 
it  d(T.  Then  da  =  dR,  and  hence  on  integrating  a  =  R  -\-  C. 
a  and  R  are  functions  of  s,  the  arc  of  the  given  curve.  Then  corre- 
sponding to  a  change  As(  =  arc  P1P2)  in  s,  cr  and  i2  will  take  on  the 
increments  Ac  and  AR  which  are  equal  by  the  foregoing  equation. 
But  A(T  =  arc  C1C2,  and  AR  =  R2  —  Ri.  Hence  arc  C1C2  equals 
^2  —  Ri- 


Fig.  114. 

157.  Involutes.  In  Fig.  114,  suppose  that  one  end  of  a  string  is 
fastened  at  C  and  that  it  is  stretched  along  the  curve  CdCiKM. 
If  now  the  string  be  unwound,  always  being  kept  taut,  the  point  M 
will,  in  accordance  with  the  properties  of  the  evolute,  trace  out  the 
curve  MP1P2P.  This  curve  is  called  the  involute  of  the  curve 
KC1C2C.  If  longer  or  shorter  lengths  of  string,  such  as  CKM2 
be  used,  other  involutes  will  be  traced.  In  fact  to  a  given  curve 
there  correspond  infinitely  many  involutes.  The  given  curve  is 
the  evolute  of  each  of  these  involutes.  We  see  that  while  a  given 
curve  has  but  one  evolute  it  has  infinitely  many  involutes. 

In  Exercise  4,  §154,  the  circle  x  =  a  cos  6,  y  =  asin  6  was  found 
as  the  evolute  of  the  curve  x  =  a(cos  6  -\-  6  sin  6),  y  =  a(sin  6  — 
6  cos  6).  Then  the  latter  curve  is  an  involute  of  the  circle.  The 
student  will  draw  a  figure  showing  a  position  of  the  string  as  it 
would  be  unwound  to  generate  the  involute  and  indicate  the 
angle  6. 


CHAPTER  XVII 

SERIES.     TAYLOR'S  AND  MACLAURIN'S  THEOREMS. 
INDETERMINATE  FORMS. 

158.  Infinite  Series.     The  expression 

Ui  +  U2-\-  U3-\-     ■    ■    ■     +  Un+     ■    '    '  (1) 

where  ui,  Uz,  "Wa,  •  •  •,  %i„,  '  •  '  is  an  unlimited  succession  of 
numbers,  is  called  an  infinite  series. 

Let  s„  denote  the  sum  of  the  first  n  terms  of  the  infinite  series 
(1).     Thus 

Sn  =  Ul  +  U2  +  U3  +    •    •   •    -\-  u„.  (2) 

If,  as  n  increases  without  limit,  Sn  approaches  a  limit  s,  this 
limit  is  called  the  sum  of  the  infinite  series,  and  the  series  is  said 
to  be  convergent. 

Illustration  1,  In  Illustration  (1),  §21,  AB,  Fig.  21,  is  a  line 
2  units  long.     The  lengths  Axi,  X1X2,  XiXz,  x^Xt,     •    •     •    x„_ix„, 

•  •  •  are  'i-,  h,  l,  h  '  '  '  >  n~i  t    '    '   '   >  respectively.     For  this 
series 

Sn    =    1    +  ^    +    ^   +   I    +     •     •     •     +     -^f  (3) 

The  limit  of  this  sum  as  n  increases  without  limit  is  2,  as  the 
figure  shows.     Or,  we  may  write, 

l  +  h  +  l  +  l+   •  •  •   +2^,+  •  ■  '    =2.  (4) 

Illustration  2.     The  sum  of  the  geometrical  progression 

1  +  r  +  r^  +  r^  +  r"  +   •  •  •   +  r"  (5) 

does  not  approach  a  limit  if  \r\  y  1,  but  if  \r\  <  1  it  approaches 

the  limit  z ,  when  n  becomes  infinite. 

1  —  r' 

The  infinite  series  (1)  is  said  to  be  divergent  or  to  diverge  if,  as  n 

increases  without  limit,  s„  does  not  approach  a  limit. 

303 


304  CALCULUS  [§158 

Thus  the  series 

l+2  +  3  +  4+---+n  +  ---, 

l-1  +  l-l  +  l-l  +  l---,    • 

1  +  R-\-  R^-\-  R^  +  R*+  •  ■  •  +  R^+  •  ■  '  {R>  I) 

are  illustrations  of  divergent  series. 

If  the  terms  of  an  infinite  series  are  functions  of  a  variable  x  and 
if  the  series  is  convergent  for  any  range  of  values  for  x,  the  series 
defines  a  function  of  x  for  that  range  of  values.  Thus,  if  |x|  <  1, 
the  series 

1  +x  +  x^  +  x^  +  •  •  •  +x^  +  •  '  '  (7) 

defines  the  function  z .     On  the  other  hand,  if    the  series  is 

1  —x  ' 

divergent  it  does  not  define  a  function  of  x.     Thus,  if  \x\  >  1,  the 

series  (7)  is  divergent  and  does  not  define  the  function  ^  _    ' 

or  any  other  function. 

It  may  happen  that  the  sum  of  a  few  terms  of  an  infinite  series 
representing  a  function  is  a  very  close  approximation  to  the  value 
of  the  function.     As  an  illustration  take  the  infinite  geometrical 

progression  (7),  which  when  |x|  <  1  represents  the  function  .  _    . 

If  the  terms  after  x*~^  are  neglected,  the  error  is 

X*  +  x*+i  +  •  •  .   +  X"  +  •  •  •   =  r^—- 

1  —  X 

X* 

The  error,  :j-^ —  is  very  small  compared  with  the  value  of  the 

function, q ,  and  decreases  as  k  increases;  i.e.,  a  better  and 

1  —  X  '        ' 

better  approximation  is  obtained  the  greater  the  number  of  terms 

retained. 

Another  infinite  series  is  obtained  by  expanding  (1  +  x)'  by 
the  binomial  theorem, 

(1  +  x)'  =  1  +  -|x  +  Ix'  -  -Ax^  +  •  •  •  (8) 

This  series  can  be  shown  to  be  convergent  when  |x|  <  1  and 
divergent  when  |x|  >  1. 

Just  as  the  function  .  _  is  represented  to  a  high  degree  of  ap- 
proximation by  the  first  few  terms  of  the  series  (7)  when  jxj  is  small, 


§159]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS        305 

the  function  {I  ■}-  x)^  is  represented  approximately  by  the  first 
few  terms  of  (8)  when  |a;|  is  small.  In  both  cases  the  functions  are 
represented  approximately  by  polynomials.  A  method  will  be 
developed  in  the  succeeding  articles  which  will  enable  us  to  deter- 
mine polynomial  approximations  to  other  functions,  such  as  sin  x, 
tan  X,  e'. 
An  infinite  series  of  the  form 

ao+  di'X  +  o.-ix'''  +  asx'  +    .    .    .    +  OnX"  +  .   .   .   (9) 
is  called  a  'power  series  in  x.     One  of  the  form 
Oo  +  ai(a;  —  a)  +  a^ix  —  ay  +  03(3;  —  a)'  +    •    •    • 

+  an{x  -  a)"  +    •    •    •  (10) 

is  called  a  power  series  in  (x  —  a).     The  series  (7)  and  (8)  are 
power  series  in  x  representing  the  functions  ^  _       and    (1  +  x)^, 

respectively.     In  the  succeeding  articles  power  series  will  be 
obtained  representing  the  functions  sin  x,  tan  x,  e*,  etc. 

159.  Rolle's  Theorem.     Let  f{x)  be  a  single-valued  continuous 
function  between  x  =  a  and  x  =  b,  having  a  continuous  first 


Fig.  116. 


Fig.  117. 


Fig.  118. 


derivative,  f'{x),  between  the  same  limits.  Further,  let  /(o)  =  0 
and  f(b)  =  0,  i.e.,  let  the  curve  representing  the  function  cross  or 
touch  the  X-axis  at  a;  =  a  and  x  =  b.     The  curve  may  or  may  not 


?o 


306 


CALCULUS 


[§160 


cross  or  touch  the  X-axis  at  intermediate  points.  (See  Fig.  115.) 
Since  f{x)  is  continuous  it  cannot  have  a  vertical  asymptote  be- 
tween X  =  a  and  x  =  b  as  shown  in  Fig.  116,  nor  can  it  have  a 
finite  discontinuity  as  shown  in  Fig.  117.  Since  f'{x)  is  con- 
tinuous between  x  =  a  and  x  =  h,  the  curve  cannot  change  its 
direction  abruptly  between  these  limits,  as  shown  in  Figs.  118  and 
119.  Since  cases  such  as  are  represented  by  Figs.  116,  117,  118, 
and  119  are  excluded,  the  curve.  Fig.  115,  must  have  a  horizontal 
tangent  at  some  point  x  =  Xi  between  x  =  a  and  x  =  b.  Hence 
the 


Fig.  119. 


Fig.  120. 


Theorem.  If  f{x)  is  a  single-valued  function  from  a;  =  a  to 
X  =  b,  andiff(x)  andf'(x)  are  continuous  between  these  limits,  and 
further  if  f{a)  =  0  andf(b)  =  0,  thenf'ixi)  =  0,  where  a  <  Xi  <  b. 

160.  Law  of  the  Mean.  Let  f(x)  be  a  single-valued  function 
between  x  =  a  and  x  =  b.  Further  let  f{x)  and  f'{x)  be  con- 
tinuous between  these  limits.  Fig.  120.  It  is  then  apparent  from 
the  figure  that  at  some  point  P  between  A  and  B,  the  tangent 
line  to  the  curve  will  be  parallel  to  the  secant  line  AB.     Hence  the 

Theorem.  If  f{x)  and  f'{x)  are  continuous  between  x  =  a  and 
X  =  b,  then 

J  ^""'^  -       b-a     ' 
where  a  <  Xi  <  b,  or 

fib)  =  /(a)  +  (&  -  a)f(x^).  (1) 

An  analytic  proof  of  this  law  will  also  be  given.     Define  a 
number  Si  by  the  equation 
fib)  =  fia)  +  ib-  a)Si,  or 

fib)  -fia)  -(6-a)>Si  =  0.  (2) 


§161]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS       307 

It  will  be  shown  that  Si  =  f'{xi),  where  a  <  Xi  <  b.  From  the 

first  member  of  (2)  build  up  the  function  0i(a;)  by  replacing  a 
by  X. 

<l>iix)  =m  -fix)  -  {b-x)Su  (3) 
Then 

<f>'i{x)  =  -f{x)+Su  (4) 

Since  f{x)  and  f'(x)  are  continuous  between  x  =  a  and  x  =  b, 
0i(a;)  and  <f>'i(x)  are  continuous  between  the  same  limits.  By  (3) 
and  (2),  <^i(a)  =  0,  and  by  (3),  <t>i{b)  =  0.  Hence  <^i(x)  satisfies 
the  conditions  of  RoUe's  Theorem  and  consequently 

<l>'i(xi)  =  0, 
or 

/'(xi)  -  5i  =  0, 
or 

Si^f'ixi), 

where  a  <  xi  <  b.     On  substituting  this  value  of  Si  in  (2)  we 
obtain 

m  =/(a)  +  (6-a)/'(xi), 

which  proves  the  theorem. 

161.  The  Extended  Law  of  the  Mean.  Let  f(x)  be  a  function 
which  with  its  first  and  second  derivatives,  f'{x)  and  /"(x),  is 
continuous  from  x  =  a  to  x  ^  b.  Define  a  number  Sz  by  the 
equation 

m  =  f{a)  +  {b-  a)f'{a)  +  ^-^-^S^,  (1) 

or 

m  -  /(a)  -  (6  -  a)f'ia)  -  ^^^^S,  =  0. 

From  the  first  member  of  the  latter  equation,  form  the  func- 
tion (f>2{x)  by  replacing  a  by  x: 

Mx)  =  m  -  /(x)  -  (6  -  x)/'(x)  -  ^^^'-5,.  (2) 

Then 

0'2(x)  =  -fix)  -  (6  -  .x)/"(x)  +/'(x)  +  (&  -  x)S, 
=  {b-x)  \S^  -  /"(x)]. 

Since /(x), /'(x),  and/"(x)  are  continuous,  ^lix)  and  4>\{x)  are 
continuous.     Further  by   (2)   and   (1),  02(a)  =  0,  and  by   (2), 


308  CALCULUS  [§162 

02(&)  =  0.     Hence  the  conditions  of  RoUe's  Theorem  are  satisfied, 
and 

<t>'2{x,)  =  0,  (3) 

where  a  <  X2  <  h. 
Or 

[h  -  X.MS2  -  fix,)]  =  0, 
or 

^2=r(x2).  (4) 

On  substituting  this  value  of  Sz  in  equation  (1),  we  obtain 

m  =  /(a)  +  (6  -  a)na)  +  ^^^^fix,),  (5) 

where  a  <  x^  <.  h. 

162.  Taylor's  Theorem  with  the  Remainder.  Finally  let  j{x) 
and  its  first  n  derivatives  be  continuous  from  x  =  a  to  x  =b. 
Define  Sn  by  the  equation 

m  =  f{a)  +  (6  -  a)r{a)  +  ^^^^f{a)  +  ■   •   • 


m  -  m  -ib-  a)f'{a)  -  ^^-—^fia)  -  •    •    • 

I  n  —  1     "^  I  n 

Form  the  function  </)„(x)  by  replacing  a  in  the  last  equation  by  x. 

*.(x)  =./(6)  -/(x)  -  (6  -  i)/'(x)  -  ^^-=^V(^)  -  ■    •    ■ 

_((^)^  (6-x)« 

n  —  1"^  _ri 

Then 

<^  "C^)  =  -      |n_i     /^"^(^)  +      |^_i     'S:,.  (3) 

Since /(x),/'(x),  •  •  •, /W(x)  are  continuous,  <^„(x)  and  </)'„(a;) 
are  continuous.  By  (2)  and  (1)  <i)n{a)  =  0,  and  by  (2)  4>n{h)  =  0. 
Hence  the  conditions  of  Rolle's  Theorem  are  satisfied,  and 

'P'niXn)    =   0, 


§162]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS        309 
or 

S„=f(-^{Xn)  (4) 

where  a  <  x„  <  6.     Hence  the 

Theorem:  If  f{x),  f{x),  f"{x),  •  •  •,  /("Ka;)  are  continuous 
from  X  =  a  to  X  =  b, 

m  =  f(a)  +  (b-  a)!' (a)  +  ^^^^  r(a)  +  ■    ■    ■ 

where  a  <.  x„  <.  b. 

This  theorem,  which  is  only  an  extension  of  the  theorem  express- 
ing the  law  of  the  mean,  is  called  Tmjlor's  Theorem  with  the 
remainder.     The  last  term  is  called  the  remainder. 

If  b  is  replaced  by  x,  (5)  becomes 

/(x)  =  /(a)  +  (X  -  a)}\a.)  +  ^"^'/'(o)  +  •    ■    - 

where  a   <   x„   <   x.      This    inequality   is    sometimes   written 
Xn  =  a  -\-  e{x  -  a),    where   0  <  B  <  I. 
Illustration  1.     Let/(x)  =  e*.     Then 

/(x)  =  e'  f{a)  =  e» 

fix)  =  e»  /'(o)  =  e» 

/"(x)  =  e'  /"(a)  =  e- 

/(")(a;)  =  e*  f^"\a)  =  e- 


Hence  by  (6) 


l  +  (x-a)+-^"-^+  •  •  •  4-(^-«)"-^ 


In  -  1 


+  (^^^,...  (7) 


If  a  =  0, 

If  a  =  0  and  x  =  1, 


«^2  /j«7l  ~"  1  ']*'* 


e=l  +  l  +  ,^+.    •    •+r^  +  T^«--  (9) 


310  CALCULUS  [§162 

The  remainder  in  (8)  and  (9)  can  be  made  as  small  as  we  please 
by  choosing  n  sufficiently  large. 

Taylor's  Theorem  may  be  expressed  in  still  another  form  by 
setting  b  in  (5)  equal  to  a  -\-  h. 

fia  +  h)=  Si.a)  +  hna)  +  ^  /"(a)  +  •  •  •  +  ^/^"^(xn)        (10) 

where  a  <  x„  <  a  +  A,  or  Xn  =  a  +  0/i,  0  <  ^  <  1. 

If  the  values  of  a  function  and  its  derivatives  are  known  at  a, 
then  the  values  of  the  function  at  a  point  a  -\-  h  can  be  computed 
by  this  formula. 

In  (10),  /(a  +  A)  is  represented  approximately  by  a  poly- 
nomial of  degree  n  —  \  in  h.  The  coefficients  are  the  derivatives 
of  /(x)  at  X  =  a.  The  error  in  the  approximation  is  given  by 
the  last  term.  This  term  gives  only  a  means  of  estimating  the 
error,  since  Xn  is  not  known.  The  maximum  error  can,  however, 
be  determined  by  substituting  M^^\  the  greatest  numerical 
value  of  /^">(a;)  in  the  interval  (a,  a -\- h)  for  f''\xr.).  The 
numerical  value  of  the  error  is  therefore  less  than 

|n 
If  a  =  0,  (6)  becomes 

f(x)  =  f(0)  +f  (0)  X  +  f"(0)  %  +  '   •  •  +  f^"~''  (0)  ^5^^ 

12  l(n-l) 

+  f^"nx»)^»      (11) 

where  0  <  Xn  <•  x,  ov  Xn  =  6x,  0  <  d  <  1. 

In  (11)  it  is  assumed  that  the  function  f{x)  and  its  first  n 
derivatives  are  continuous  from  x  =  0  to  x  =  a.  (11)  is  known 
as  Maclaurin's  Theorem  with  the  remainder. 

Illustration  2.  Expand  sin  x  by  Maclaurin's  Theorem  in  powers 
of  z  as  far  as  the  term  containing  x®. 


/(x)  =        sin  X 

/(O) 

= 

0 

f{x)  =        cos  X 

/'(O) 

= 

1 

/"(x)  =  -  sin  X 

/"(O) 

= 

0 

fix)  =  -  cos  X 

/'"(O) 

= 

-  1 

/■^(x)  =       sin  X 

rxo) 

= 

0 

P{x)  =        cos  X 

/"(O) 

= 

1 

/^*(x)  =  —  sin  X 

rm 

= 

0 

/^"(x)  =  -  cos  X 

r'\x.) 

= 

—  COS  X7. 

§163]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS       311 
Substitution  in  (11)  gives 


X'' 


sina;  =  x  —  ro+j-g—  i^  cos  Xi,  (12) 


where  0  <.  Xt  <.  x. 

Since  |cos  X7I  <  1,  sin  x  differs  from 
x^       x^ 

^~  [3  +  j5 

x'' 
by  a  number  less  than  r«* 

In  general,  since  sin  x  and  its  derivatives  are  continuous, 

x^    ,   x^  x"  ,  ,  ^      ,     s 

smx  =  X  —  r^  +  re  ~  '    '    "if"  (sin  x„  or  cos  x,,),    (13) 

where  0  <  x„  <.  x.     Thus  the  difference  between  sin  x  and 

^  ~  [3  +  [5   ~  ■    ■    "  -  \r^^^ 

x" 
is  less  than  r—,  a  number  which  for  a  given  x  can  be  made  as  small 
\n 

as  we  please  by  taking  n  sufficiently  large.  Hence  the  series  (13) 
can  be  used  in  computing  the  value  of  sin  x.  If  x  is  small,  only  a 
few  terms  of  the  series  need  be  used  to  obtain  a  very  close  ap- 
proximation to  sin  X.  Thus  in  formulas  in  which  sin  x  occurs, 
sin  X  is  frequently  replaced  by  x  if  the  angle  is  small.  Such  a  sub- 
stitution was  made  in  equation  1,  §81.  It  must  be  remembered 
in  making  the  substitution  that  x  is  expressed  in  radians. 

163.  Taylor's  and  Maclaurin's  Series.  If  f{x)  and  all  of  its 
derivatives  are  continuous  within  an  interval,  the  number  of 
terms  in  (6),  (10),  and  (11),  §162,  can  be  increased  indefinitely. 
These  equations  then  become,  respectively, 

f  (x)  =  f(a)  +  f  (a)  (X  -  a)  +  f"(a)  ^^^'  +  •   •  • 

+  f^"Ha)^^^%---.  (1) 

f(a+h)=f(a)  +  f'(a)h  +  r(a)j'^+-    •    •  +  f (">(a)^+ •  •  •.   (2) 

f(x)=f(0)  +f  (0)x+  f"(0)^^  +  •    •    •  +  f  ^"^  (0)^  +  •  •  ••  (3) 


312  CALCULUS  [§163 

In  (1),  f{x)  and  its  derivatives  are  assumed  to  be  continuous 
from  a  to  x. 

In  (2) ,  /(x)  and  its  derivatives  are  assumed  to  be  continuous  from 
aio  a  +  h. 

In  (3),  /(x)  and  its  derivatives  are  assumed  to  be  continuous 
from  0  to  X. 

The  series  (1)  and  (2)  are  called  Taylor's  Series  and  (3)  is  called 
Maclaurin's  Series. 

If  we  denote  the  last  term  in  each  of  the  equations  (6),  (10),  and 
(11),  §162,  by  Rn,  it  is  necessary  that 

i^h^n    =     0 

in  order  that  (1),  (2),  and  (3)  shall  represent /(x),  /(o  +  h),  and 
/(x),  respectively. 

Such  series  represent  a  function  only  so  long  as  they  are  conver- 
gent. Later  in  this  chapter  means  of  testing  the  convergence  of 
series  will  be  discussed.  The  series  (1),  (2),  and  (3),  if  convergent, 
represent  /(x)  but  do  not  give  a  means  of  estimating  the  error 
made  by  stopping  with  a  given  term.  This  can  best  be  deter- 
mined from  the  expression  for  the  remainder  72„  in  Taylor's  or 
Maclaurin's  Theorem  with  the  remainder. 

Illustration  1.  Represent  sin  x  by  a  power  series  in  (x  —  a). 
Use  formula  (1). 

/(x)  =  sin  X  /(a)  =  sin  a 

/'(x)  =  cos  X  f'ia)  =  cos  a 

j"  (x)  =  —  sin  X  /"(o)  =  —  sin  a. 

/'"(x)  =  —  cos  X  /'"(«)  =  —  cos  a 

fix)  =  sin  X  /'^(«)  =  sin  a 

P{x)  =  cos  X  /^(o)  =  cos  a 

Then  by  (1) 

,              ,           .         .        (x-  ay  (x  -  ay 

sin  X  =  sin  a  +  cos  a  {x  —  n)  —  sm  a  — r^ cos  a  — r^ — 

.        (x  -  a)*  {x  -ay 

+  sm  a  — Hi +  cos  a      ,  - — -   —   •  •  •. 

The  corresponding  Maclaurin's  Series  is  obtained  by  letting  a  =  0. 

x'    ,    x^        x^    , 
sin  X  =  x  -  rg   +  ,  5  -  1^  +  •  .  •. 


§164]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS         313 

Illustration  2.     Expand  tan  x  in  a  power  series  in  x. 

fix)  =  tan  X.  /(O)  =  0 

fix)  =  sec^x.  f'iO)  =  1 

fix)  =  2  sec^x  tan  x.  /"(O)  =  0 

=  2(tana;  +  tan'x). 

fix)  =  2(sec2a;  +  3  tan^x  sec^a;)  /'"(O)  =  2 

=  2(1  +  4  tan^x  +  3  tan%) 

fix)  =  16  tan  X  sec^x  +  24  tan^x  sec^x  /"'(O)  =  0 

=  16  tan  X  +  40  tan^x  +  24  tan^x 

fix)  =  16  sec^x  +  120  tan^x  sec^x  +  120  tan^x  sec^x  fiO)  =  16 

On  substituting  in  (3)  we  obtain 

X'       2x« 
tan  x  =  x  +  -^  +  Yk''t~''  '• 

The  next  two  terms  are  ^W  ^^  and  ^H^  x". 

Exercises 

1.  Expand  cos  x  in  a  power  series  in  x. 

2.  Expand  cos  x  in  a  power  series  in  (x  —  a). 

3.  Expand  cos  (a+  ^0  in  a  power  series  in  h. 

4.  Expand  sin  (a  +  A)  in  a  power  series  in  h. 

6.  Express  the  remainder  after  three  terms  in  each  of  the  series 
of  Exercises  1,  2,  3,  4. 

6.  Expand  e"  in  a  power  series  in  x. 

7.  Expand  e"'^'^  in  a  power  series  in  h. 

8.  Expand  e"  in  a  power  series  in  {x  —  a). 

9.  Expand  log  (1  +  x)  in  a  power  series  in  x. 

10.  Expand  log  (1  —  x)  in  a  power  series  in  x. 

11.  Expand  tan'^x  in  a  power  series  in  x. 

12.  By  the  use  of  the  series  already  found,  compute: 
(a)   ^e  to  5  decimal  places. 

(6)    \^e  to  6  decimal  places, 

(c)  sin  3°  to  6  decimal  places. 

(d)  cosine  of  1  radian  to  4  decimal  places. 

13.  By  the  use  of  the  result  of  Exercise  3,  find  cos  33°  correct  to  4 
decimal  places. 

14.  By  the  use  of  the  result  of  Exercise  4,  find  sin  32°  correct  to  4 
decimal  places. 

164.  Second  Proof  for  Taylor's  and  Maclaurin's  Series.     These 


314  .  CALCULUS  [§164 

series  can  be  obtained  very  simply  in  another  way  if  we  make 
certain  assumptions  and  do  not  attempt  to  justify  them. 

Assume  that  j{x)  can  be  represented  by  an  infinite  power  series 
in  (x  —  a): 

f{x)  =  aa-\-ai{x—a)-\-a2{x—aY+'  •  •  +a„(a;— a)"+ •  •  •,  (1) 

where  Oo,  Oi,  02,  '  '  ',  On,  '  '  '  are  coefficients  which  are  to  be 
determined.  Assume  further  that  the  result  of  differentiating  the 
second  member  term  by  term  any  given  number  of  times,  is  equal 
to  the  corresponding  derivative  of  the  first  member.     Then, 

f'{x)  =  ai  +  2a2(x  —  a)  -\-  803(0;  —  a)^ 

+  •  •  •  +  nan(x  —  a)"~^  +  •  ■ 
fix)  =  202  +  Ga^ix  -  a) 

+  ■  '  ■+  n{n  -  l)a„(a;  -  a)--^  +  ■  ■  ■    (2) 
f"'ix)  =  603  +  •  .  •  +  n(n  -  l)(n  -  2)a„(x  -  a)"-^  +•  • 


/(")  (x)  =  |wo„  +  .  .  . 
Put  re  =  a  in  (1)  and  (2). 


whence 


/(a)  =  ao, 
/'(a)  =  ai, 
r(a)  =  2a2, 
/'"(a)  =  I3a3, 

/"Ha)  =  Inan. 

ao  =  /(a), 
ai  =  f'(a), 

fia) 
a2  =  ^v 

at  =  —f^ — > 


/("^  («) 

Ctn    =    , > 


§165]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS       315 

Substituting  in  (1)  we  obtain 

fix)  =  f{a)  +  /'(a)(x  -a)+  -^(x  -  a)' 

By  setting  a  =  0,  and  x  =  a  +  h,we  get  (3)  and  (2),  respectively, 
of  the  preceding  section. 

165.  Tests  for  the  Convergence  of  Series.  Several  tests  will 
now  be  given  for  determining  whether  or  not  a  series  is  convergent. 
They  will  be  given  without  proof,  though  in  most  cases  the  proof 
is  not  difficult. 

If  a  series  Ui  -\-  uz  -{•  •  •  •  +  Wn  +  •  •  -is  convergent, 

lina    Un  =  0. 
n—  00 

The  converse  of  this  statement  is  not  true.     Thus  the  series 

1  +  i  +  i  +  i  +  i  +  •  •  •  +  ^  +  •  •  •  (1) 


is  divergent,  although 


^^"^  -  =  0. 

n=oo^ 


That  this  series  is  divergent  can  easily  be  seen  as  follows: 

3    T^    4    -^    2 

"5   +    C    +  Y  +  "8    >    2 

«    ~r   1  0'   +     1  f   -P   "1  2"   "T    1  3'   +   "1  4"   "T    I  5    "1"     1  6"    ^    2 

The  terms  of  the  series  can  then  be  grouped  into  infinitely  many 
groups  such  that  the  sum  of  the  terms  in  each  group  is  greater 
than  5.     But  the  series 

is  divergent.     Much  more  then  is  the  series  (1)  divergent. 

Test  1.     If  ^^"^  Un  is  not  zero  the  series  is  divergent.     This  test 

n=  <x> 

is  easy  to  apply  and  if  it  shows  the  series  to  be  divergent,  no 
further  investigation  is  necessary. 


316  CALCULUS  f§166 

Test  2.  Alternating  series.  A  series  of  decreasing  terms  whose 
signs  are  alternately  plus  and  minus  and  for  which 

n—  00 

is  convergent. 
Thus  the  series 

l-l  +  l-\  +  l-ls+  •  •  '  (2) 

is  convergent. 

The  reason  for  the  convergence  of  such  an  alternating  series 
can  be  seen  as  follows.  Denote  by  Sn  the  sum  of  the  first  n 
terms  and  suppose  the  (n  +  1)**^  term  positive.  (See  Fig.  12L) 
Then,  since  the  terms  are  constantly  decreasing, 

Sn  +  l   >    Sn',    Sn+2    <^    Sn  +  l',    Sn+2   >   Sn- 


'Sfl+l ' 


'\^ 


Sn+2  -*{ 

Fig.  121. 

It  is  clear  that  as  n  increases  Sn  oscillates  back  and  forth  but 
always  within  narrower  and  narrower  limits,  owing  to  the  fact 
that  the  terms  are  constantly  decreasing.  As  n  becomes  infinite 
the  amount  of  this  oscillation  approaches  zero  since 

71=  00 

Sn  therefore  approaches  a  limit. 

Test  3.  Comparison  Test.  If  the  terms  of  a  series  are  in  numer- 
ical value  less  than  or  equal  to  the  corresponding  terms  of  a  known 
convergent  series  of  positive  terms,  the  series  is  convergent.  If  the 
terms  of  a  series  of  positive  terms  are  greater  than  or  equal  to  the 
corresponding  terms  of  a  divergent  series  of  positive  terms,  the  series 
is  divergent. 

A  useful  series  for  comparison  is  the  geometrical  series 

a  -\-  ar  -^  ar^  +  ar''  +  ■  •  •   +  ar"  +    *  *  •  ,  (3) 

which  is  convergent  if  |r|  <  1  and  divergent  if  |r|  ^  L  See  also  the 
series  (a)  of  Illustration  2  of  this  section. 


§165]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS        317 

Test  4.     The  Ratio  Test.     By  comparison  with  the  geometrical 
series  it  can  be  shown  that  the  series 


is  convergent  if 
divergent  if 
If 


lim 

Un+l 

n=a> 

Un 

lim 

Un+l 

n=co 

Un 

lim 

Un+l 

«=« 

Un 

<1, 


>  1. 


=  1 


the  test  fails.     In  this  case  other  tests  must  be  appHed. 

There  are  a  great  many  tests  for  the  convergence  of  series  but 
only  a  few  can  be  given  here.  It  should  be  added  that  there  is  no 
test  that  can  be  applied  to  all  cases. 

Illustration  1.     Test  the  series 


1  -  t+l-i  + 


+ 


for  convergence. 
Since 


lii 


Un 


is  not  zero,  the  series  is  divergent  (Test  1).     It  is  to  be  noted  that 
the  terms  of  the  series  are  alternately  positive  and  negative  and 
that  they  decrease,  but  they  decrease  to  the  limiting  value  1 
instead  of  0.     Hence  test  2  does  not  apply. 
Illustration  2.     Test  the  series 


1  +  -+-+-"  + 

^  ~  9<    '    *?«  ~  /!<     ' 


(a) 


for  convergence.     This  series  is  useful  in  testing  the  convergence  of 
series  by  comparison. 

If  f  =  1  we  have  seen  that  this  series  is  divergent.  (See  (1).) 
If  f  <  1  each  term  of  (a)  is  greater  than  the  corresponding  term 
of  (1)  and  hence  (o)  is  divergent.  If  f  >  1  we  can  compare  (a) 
with 

l  +  l  +  l  +  l,  +  i  +  ll  +  l.+  "•  (« 


318 


CALCULUS 


r§165 


Each  term  of  (a)  is  less  than  or  equal  to  the  corresponding  term 
of  (6).     But  (6)  is  convergent  since  it  can  be  written 


^  +  ^(^)+M4^)+M^)-^ 


1+-+-+^+-+ 


which  is  a  geometric  series  whose  ratio,  ^~,  is  less  than  1.     Hence 

(a)  is  convergent  when  t  >  1.    Summing  up: 
(a)  is  divergent  if  t  ^  1. 
(a)  is  convergent  U  t  >  1. 

Illustration  3.     Test  the  series 


1 


1 


i  +  r2  +  r3+i4  +  -  •  • 

for  convergence. 

Apply  test  4. 

1 

Mn    =    P- 

jn 

Hence 

1 

lim 

Un+1 

lim  1^  +  1        lim       1 

M=oo 

Un 

1             "=-  n  +  1 

In 

=  0. 


The  series  is  therefore  convergent. 

Illiistration  4.     For  what  values  of  x,  if  any,  is  the  series 


^~  13"^  15       17  + 


convergent? 


Then 


2n  -1 


lim 

n=  00 


Un+l 


Un 


a;2„+i 

im 

|2n  +  l 

=  QO 

a;2„-i 

|2n-l 

lim 

n  =  oo 


|2n  (2n  +  1) 


=  0, 


§165]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS       319 

for  all  finite  values  of  x.     Hence  the  series  is  convergent  for  all 
finite  values  of  x,  positive  or  negative. 
Illustration  5.     For  what  values  of  x  is 


X'' 

'^3 


X*   ,x. 


convergent? 


\Un\    = 

71 

_    lim 

^n  +  l 

= 

lim 

n=  00 

X 

Un  +  l 

n+  1 

n 

Un 

n 

+  1 

n 

lim 

n=oo 


The  series  is  therefore  convergent  if  |x|  <  1.  Furthermore  it  is 
convergent  if  x  =  1  (Test  2),  and  divergent  if  x  =  —  1.  (See 
series  (1).) 

As  has  been  stated  there  is  no  one  test  of  convergence  which 
can  be  applied  with  certainty  of  success  to  any  given  series.  The 
tests  which  can  be  most  frequently  applied  have  been  given.  It 
is  suggested  that  the  following  procedure  be  observed  in  general. 

1.  See  if  ^^^Un  =  0. 

n=  00 

2.  If  so,  is  test  2  applicable? 

3.  Tf  not,  try  the  fatio  test,  test  4.     This  will  fail  if 


lim 

n=  00 


M„ 


1. 


4.  In  this  case,  and  in  cases  where  the  other  tests  fail  or  are 
difficult,  try  the  comparison  test. 


Exercises 

Test  the  following  scries  for  convergence. 

\.  \  -\+\  -I  +  i\  -■  •  •■ 

2.  i  -  I  +  -f  -  I  +  A  -  •  •  •• 

3.  4  +  i  +  ^  +  ^  +  A  +  •  •   •. 
|3        |4         |5 

10*  ^ 

^     A      A 

13  +  14+  I  5  +  '   '   *• 


5. 


10  "*"  10^  "^  10' 


1 

12  + 


320  CALCULUS  [§166 


^-  ^   +  2V2  +  3V3  +4V4  +  "  '  ■• 
'  32  ^  52       72  ^9=* 

^'  1-2  ^  3-4  ^  5-6  ^ 
For  what  values  of  x  are  the  following  series  convergent? 

10.  The  Maclaurin's  series  for  e"?     Exercise  6,  §163. 

11.  The  Maclaurin's  series  for  cos  x?     Exercise  1,  §163. 

12.  The  Maclaurin's  series  for  sin  x?     Illustration  1,  §163. 

13.  The  Maclaurin's  series  for  log  (1  —  x)?    Exercise  10,  §163. 

14.  The  Maclaurin's  series  for  tan~^x?     Exercise  11,  §163. 

166.  Computation  of  Logarithms.  The  series  of  Exercise  9, 
§163,  for  log  (1  +  x)  is  convergent  only  when  —1  <x^  +  1,  and 
that  for  log  (1  —  x),  Exercise  10,  §163,  only  when  —  1  ^x<  +  1. 
It  would  appear  then  impossible  to  find  the  logarithm  of  a  number 
greater  than  2  by  these  formulas.  By  a  very  simple  device  it  is, 
however,  possible  to  obtain  formulas  for  finding  the  logarithm 
of  any  number. 

From  the  series  of  Exercises  9  and  10,  §163  it  follows  that 

1  -f  J 
log:j =  log  (1  +  x)  -  log  (1  -  x) 


X+3-+g-+---],  (1) 


where  Ixl  <  1.     Let  x  =  ^ — i~T'     Then 
'   '  2z  +  1 

1  -hx  _  z+  1 

1  —  X  z 

z 
where  0  >  0,  or 

log(^+l)=log3+2[2^+3(22  +  l)3+5(22Vl)^+'  '^^^ 

By  letting  z  =  1,  log  2  can  be  computed  by  this  formula.  The 
series  is  much  more  rapidly  convergent  than  that  for  log  (1  -\-  x), 
X  =  1.     In  fact,  100  terms  of  the  latter  series  must  be  taken  to 


and 

^°^  ~T~  ^  ^L2M^  "^  3{2z  +  iy  "^  5(27+Tr^  +  •  •  -J .   (2) 


§168]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS       321 

obtain  log  2  correct  to  two  decimal  places,  while  four  terms  of  the 
new  series  (3)  will  give  log  2  correct  to  four  decimal  places.  After 
log  2  has  been  found,  log  3  can  be  found  by  setting  z  =  2.  The 
logarithm  of  4  is  found  by  taking  twice  log  2;  log  5  by  setting 
2  =  4;  log  6  by  adding  log  3  and  log  2,  and  so  on. 

Exercise 
Compute  log  5  correct  to  four  decimal  places,   given  that  log  4 
=  1.38629.     Here,  as  always  in  the  Calculus,  the  base  is  understood 
to  be  e. 

167.  Computation  of  tt.  By  letting  a;  =  1  in  the  series  for 
tan~'x.  Exercise  11,  §163,  the  following  equation  is  obtained  from 
which  IT  can  be  computed: 

I  =  tan-i  l  =  l-i  +  i-|+-    •    •• 

This  series  converges  very  slowly.  To  obtain  a  more  rapidly  con- 
verging series  make  use  of  the  relation 

tan~^  1  =  tan"^  ^  +  tan~^  |. 
Then 

4       '       (3)  (23)  +  (5)  (2^)       (7)  (2^)^ 

^'       (3)(33)  ^  (5)(3^)       (7)(3^)^ 

168.  Relation  between  the  Exponential  and  Circular  Functions. 
If  it  be  admitted  that  the  Maclaurin's  series  expansion 

e'  =  l+2  +  |2  +  ^  +  -  •  •,  (1) 

which  was  proved  for  real  values  of  z,  is  also  true  when  z  is  imagin- 
ary, we  obtain,  on  setting  z  =  ix, 

6.^  =  l-hta;+^-h-|3--|-^-h-|5-  +  ^  +  -|7-+-  ■  • 

^    ,    ,        x^      ix^   ,  X*  ,  ix^      x^      ix''   , 

=  l+tx-j2-  ||  +  |4+^-j6--|7  +  -  •  •  (2) 

On  separating  real  and  imaginary  parts  this  becomes 

e"-l-^  +  |4-|g-|---- 


+  1(2 


13    '   15       17 
21 


+  %-%+■■).  (3) 


322 


CALCULUS 


[§168 


Since  (Exercise  1  and  Illustration  1,  §163), 


and 


cosx  =  l-j2-  +  |^-ig  + 


x^    ,    x^       x*   . 
sinx=x-j3+^-i^  + 


it  follows  that 

e«  =  cos  X  +  i  sin  X. 

On  changing  the  sign  of  x  it  results  that 

e-»^  =  cos  X  —  1  sin  X. 

Solving  equations  (4)  and  (5)  for  cos  x  and  sin  x, 


and 


cosx  = 


sinx  = 


2 

e**  —  e- 


(4) 

(5) 

(6) 
(7) 


These  interesting  relations  between  the  circular  and  exponential 
functions  are  of  very  great  importance. 


Fig.  122. 


Fig.  123. 


If  Q  represent  the  vectorial  angle  in  the  complex  number  plane, 
then  it  is  clear  from  Fig.  122  that  e*'  represents  a  point  on  the 
unit  circle  (circle  of  radius  1  about  the  origin  as  center)  in  this 
plane.  Further,  any  complex  number  a  +  hi  can  be  put  in  the 
form  pe".     For  (Fig.  123) 

a  -^-hi  =  p  (cos  0  +  1  sin  Q)  =  pe*", 

where  p  =  y/a^  +  h"^. 


5.  e*"-. 

7. 

e^*'. 

6.  e-'"". 

8. 

3tV 

5e  4  . 

§169]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS       323 

Exercises 

Represent  by  a  point  in  the  complex  plane : 

IX  IT 

1.  3e^  3.  e^. 

2.  2e"  3  ,  4.  e^. 
9.  Express  the  numbers  of  Exercises  1-8  in  the  form  a  +  hi. 

169.  DeMoivre's  Theorem.     The   interesting  and  important 
theorem,  known  as  DeMoivre's  Theorem, 

(cos  0  +  i  sin  0)"  =  cos  n0  +  i  sin  n^  (1) 

can  be  easily  established  by  the  use  of  the  relation  (4)  of  §168. 

For, 

(cos  d  +  i  sin  0)»  =  {e'^Y  =  e*"^  =  cos  nd  +  i  sin  nd. 

Exercises 

Find,  by  the  use  of  (1), 

1.  The  cube  of  1  +  i.  1  +  iy/^ 

4.   Ihe  cube  of 


2.  The  square  of ^ •  -    n^i  i        -.  —  1  —  iy/Z 

2  5.  The  cube  of ^ -• 

3.  The  cube  of ^ 6.  The  cube  of ^ •• 

In  (1),  11  may  be  a  fraction  as  well  as  an  integer.     It  will  then 

indicate  a  root  instead  of  a  power.     In  this  case  we  do  not  have 

simply  one  root: 

—  fl  ft 

(cos  d  -\-  i  sin  0)'"  =  cos [-  i  sin—  > 

mm 

(n  having  been  placed  equal  to  — ,  where  m  is  an  integer)  but 
m  —  1  additional  roots.     This  follows  from  the  fact  that 

giO   _   gi(0  +  2p»)  (2) 

where  p  =  0,  1,  2,  3,  4,  •  •  ■  ,m,m  -\-  \,  •  •  • .     Hence  we  can  write 

(cos  d  +  I  sin  ^)™  =  [e^"]'"  =  [e'^"  +  ^p^^^^ 
or 

1^  t(0  +  2pw) 

(cos  ^  +  t  sin  d)""  =  e       »»       ,  (p  =  0,  1,  2,  •  •  •)  ^^' 


324  CALCULUS  [§170 

It  would  appear  at  first  sight  as  if  there  were  infinitely  many 
roots  corresponding  to  the  infinitely  many  values  of  p.  But  a 
little  consideration  shows  that  when  p  y  m,  the  roots  already 
found  by  letting  p  take  the  values  0,  1,  2,  •  •  • ,  w  —  1,  repeat 
themselves,  since  e^"  =  1.  There  are  then  exactly  m  mM^  roots 
of  e*^  =  cos  0  +  I  sin  d, 

e- —^  =  cos  — ' — —  +  t  sm  — ^ — —,  (4) 

where  p  =  0,1,2,  •  •  • ,  w  —  1. 

Illustration.     Find  the  three  cube  roots  of  —  1. 

(-l)i  =  (e-)i 

^  ^gtu  +  2px)j}  (p  =  0,1,2) 

t(ir  +  2px) 

=  e      3  (p  =  0,  1,  2) 

tV  5jt 

=  e^ ,    e*',      and  c  ^  . 

Exercises 

ie 

1.  Show  that  the  three  cube  roots  of  a  +  6t  =  pe'*  are :     -y^  «  ^i 
«(g  +  2t)  i(g  -f  4t) 

■^^  e  3  J  and  -^^  c  3  .  How  would  these  roots  be  deter- 
mined graphically? 

2.  Find  the  two  square  roots  of  1  +  i. 

3.  Find  graphically  the  two  square  roots  of  l. 

4.  Find  graphically  the  three  cube  roots  of  1. 

170.  Indeterminate  Forms.     It  has  already  been  shown  that 

^  has  no  meaning.     See  §25.     Thus 

x-2 
has  no  meaning  at  a;  =  2.     Its  value  at  x  =  2  is  defined  as 

lim   ^^  ~  '^  =  4 

x=2    X  —  2 

Similarly 

sin  a 
tan  a 
has  no  meaning  at  a  =  0.     Its  value  at  a  =  0  is  defined  as 
lim  sin  a  ^  ^ 
°=o  tana 


§170]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS        325 


In  general,  if  </)(a)  =  0  and  /(a)  =  0,  the  value  of  the  function 
lim  ^xji 

x  =  a  f(^x)  ' 


-77-v  Sii  X  =  a  IS  denned  as 


The  calculation  of  this  limit  is  simplified  in  many  cases  by  the 
application  of  the  law  of  the  mean.     See  §160. 

Thus,  let  it  be  given  that  4>{a)  =0  and  /(a)  =  0  and  let  4)(x) 
and  f(x)  satisfy  the  conditions  imposed  in  the  statement  of  the 
law  of  the  mean.     Then 

lim  0(a;)  ^   Hm  </>(»)  +  (x  -  a)  4>'  [a  +  0\{x  -  a)] 
x^a  /(^)         x^a  ji^a)  +  {x-a)  f  [a  +  d^ix  -  a)] 

__   lim  4>'[a  +  ei(x-  a)]       0^(a)      ^    .^i  .  . 
-    z^a  f'[a  +  e,{x-a)]        /'(a) '     "^  ^6,^  ^^ 

If  (f>'(a)  and /'(a)  are  also  zero,  we  make  use  of  the  extended  law 
of  the  mean.     Thus 

0(a)  +  (x  -  a)  0'(a)  +  ^^-^V"[a  +  ^1(0;- a)l 
lim  4>{x)  ^  lim |i 

•'^  ^  /(«)  +  (x-  a)  f'(a)  +  ^-^—^ria  +  d^ix  -  a)] 

_  lim  </>'[a  +  gi  (x  -  g)]       <i>"(a) 
x^a  j"[a  +  d,{x  -  a)\        f{a)  ' 

The  process  is  to  be  continued  further  if  f"(a)  and  <^"(a)  are 
both  zero. 

Illustration  1. 

lim  e'  -1  ^ 

x=o        X 

Illustration  2. 

lim  e^  —  e~'  —  2x 


a:=&0 


X  —  sm  a: 


Hm  e- 

z=0     1 

=  1. 

lim  € 

.  +  e-x_ 

2 

'    x=0 

1  —  cos  X 

lim  e^ 

.  _  g-x 

x=0 

sin  X 

lim  e' 

+  e-* 

1=0 

p.na  X 

2. 


326  CALCULUS  [§170 

The  Form  — .     The  same  process  is  employed  in  evaluating 

00 

the  indeterminate  form  — -.     The  proof  is  omitted. 

Illustration  3. 

lim  ^  _  lim  2x  _  lim  2^  _« 

X—  00    nx  X—  oo   gx  X=  00   ox  ' 

The  Form  0  <» .     The  indeterminate  form  0  «>   can  be  thrown 

0  00 

into  either  of  the  forms  -  or  — ■.     Thus 

0  00 

lim  3.  „„.  y  _  lim  _^ lim      1      _ . 

=^=0  -^  ^"^  ^       ^=0  tan  X       ^^0  sec2  x~ 

Other  indeterminate  forms  are :    °o  —  «> ,  1  °°,  0",  °°°. 
Thus,  if  <p(x)  and  /(x)  become  infinite  for  x  ==  a,  (/>(a)  —  /(a)  is 
defined  as 

^^^JHx)  -  fix)]. 

This  expression  can  be  written 

1  1 

lim  r ,  /  V       ,/  ^^  _  lim  fix)       <t>{x) 

W)f(x) 
an  indeterminate  form  of  the  type  ^. 

If  (l>{x)  becomes  infinite  and/(x)  becomes  1  for  x  =  a,  [/(a)]*^"^ 
is  defined  as 

This  limit  can  be  calculated  as  follows.     Let  y  =  [/(x)]*^^'. 
Then 

logy  =  4){x)  log/(x)  = ^ > 

W) 
an  indeterminate  form  of  the  type  -.    If 

lim  log /(a;) 

W) 


§170]        TAYLOR'S  AND  MACLAURIN'S  THEOREMS       327 
is  found  to  be  c,  then 


lim  „  =  g« 

x  =  a 


The  two  remaining  forms  are  evaluated  in  a  manner  similar  to 
the  last. 

Many  indeterminate  forms  can  be  evaluated  directly  by  simple 
algebraic  transformations. 


Exercises 


Evaluate  the  following: 

lim   loga^  13     lim  _^. 

J-  x=l  X  -  i  •  2=°'  log  a; 


lim    1  -cosg  14.    lim  ^ 

lim  c» 


9=0  cosflsin^fl 


«    lim  a;  COS  a;  —  sin  x  16.  ^^  „  ^ 


X  =  0  X 

4. 


lim  tan  x  -  sin  x  16.  ^™  e«  tan  -• 

z=0      a;  —  sin  X 

lim  a:"  -  1 


^-  a;=l  X  -  1  ^„    lim 


lim  sin  33; 
•  1=0  sin  2a; 


17    lim  r_^ L_" 

^''    X=l    Lx2    -    1  X-l_ 

i™    [(|-9)ta„; 

2 

-    i=u  smzx  r       r    1  11 

^    lim  tan  3a;  19.  ^^^  |^j^  -  a._ij' 

g    Um       ^^  20.1;-  (cos  xfot^. 
x=o  sin  X  —  a;  1 

lim  x^  -  X  -  6  21.  iTod-^)^- 
^-  x=3      x2  —  9  i:,„ 

,„,ta  311+5  22.'-(-x).»«. 

•  I*  .  4i!  +  1  jj_  hm  (3j„  ^)t.n  . 
lim   log  X 

12.    li™   '^-  „.    lim  tan  fl. 


CHAPTER  XVIII 

TOTAL  DERIVATIVE.    EXACT  DIFFERENTIAL 

171.  The  Total  Derivative.  Let  z  =  f{x,  y)  and  let  x  and  y  be 
functions  of  a  third  variable  t,  the  time  for  example.     We  seek  an 

dz 
expression  for   -rr,  the  derivative  of  z  with  respect  to  t,  in  terms  of 

dx      J  dy 

As  an  illustration  of  what  is  meant,  let  z  denote  the  area  of  a 
rectangle  whose  sides  x  and  y  are  functions  of  t,  and  at  a  given 
instant  let  each  side  be  changing  at  a  certain  rate.  The  rate  at 
which  the  area  is  changing  is  sought. 

Returning  to  the  general  problem  let  t  take  on  an  increment  At. 
Then  x  takes  on  the  increment  Ax  and  y  the  increment  Ay,  and 
consequently  z  the  increment  Az.     We  then  have 

2  =  fix,  y)  (1) 

2  -f  Az  =  /(x  +  Ax,  2/  +  Ay) 

Az=J{z  +  Ax,y  +  Ay)  -  /(x,  y)  (2) 

Az  =  /(x+Ax,  y+Ay)-J{x,  y+Ay)+f{x,  y+Ay)-f(x,  y)  (3) 
.     Az  ^  fix  -\-Ax,y  +  Ay)  -  /(x,  y  +  Ay)  Ax 
At  Ax  At 

fix,  y  +  Ay)  -  fix,  y)  Ay 
"•"  Ay  At        ^  ' 

Taking  the  limits  of  both  sides  of  (4)  as  At  approaches  zero,  we 
have 

dz  _  dfjx,  y)  dx       dfjx,  y)  dy  ,  . 

dt  dx      dt  "^      dy      dt'  ^  ' 

since  Ax  and  Ay  each  approach  zero  as  At  approaches  zero. 
Equation  (5)  can  be  written  in  the  form 

dz  _  3z  dx       3z  dy  ,  . 

dt  ~  ax  dt  "•"  ay  dt  *  ^^ 

328 


§171]    TOTAL  DERIVATIVE.     EXACT  DIFFERENTIAL    329 

This  states  that  the  rate  of  change  of  z  with  respect  to  t  is  equal 
to  the  rate  of  change  of  z  with  respect  to  x,  times  the  rate  of  change 
of  X  with  respect  to  f,  plus  the  rate  of  change  of  z  with  respect  to 
y,  times  the  rate  of  change  of  y  with  respect  to  i. 
\il  =  Xy  (6)  becomes 

dz 

dx 


dz       dz   dy 
dx       dy  dx 


This  formula  applies  when  z  =  f{x,  y)  and  ?/  is  a  function  of  x, 
e.g.,  y  =  <i>{x). 

Multiplying  (6)  by  dt  we  obtain 


dz 


^r  dx  +  ^-  dy. 
dx  dy 


(7) 


This  defines  dz,  which  is  called  the  total  differential  of  z. 


z 

E 

I 

S 

c 

F 
R 
A 

f 

K 

% 

7 

Y 

dy 

r 

/ 

Fig.  124. 

We  shall  now  give  a  geometrical  interpretation  of  dz.     Let 
P,  Fig.  124,  be  the  point  {x,  y,  z)  on  the  surface  z  =  fix,  y). 
Let 

PC  =  dy 
and 

PA  =  dx. 

Then  Q  is  the  point  {x  +  dx,  y  +  dy,  z  +  Az).  Let  PDEF  be 
the  plane  tangent  to  the  surface  at  the  point  P.  Then  PF  is 
tangent  to  the  arc  PR,  and  PD  is  tangent  to  the  arc  PS. 


330  CALCULUS  [§171 

From  F  draw  FK  parallel  to  AB  meeting  BE  in  K. 

BE  =  BK-\-  KE 

BK  =  AF  =  ^  dx. 
ox 

Since  FK  =  PC  and  PD  =  FE,  triangle  KFE  is  equal  to  the 
triangle  CPD,  and 

KE  ^  CD  =  ^  dy. 
dy    ^ 

Therefore 

^^  =  dx^  +  3y^y' 

Hence  BE  =  dz.  Consequently  dz  may  be  interpreted  as  the 
increment  measured  to  the  tangent  plane  to  2  =  f{x,  y)  at  the  point 
P  (x,  y,  z)  when  x  and  y  are  given  the  increments  dx  and  dy 
respectively. 

Illustration  1.     \i  z  =  xy,  the  area  of  a  rectangle  of  sides  x 
and  y,  we  obtain  by  using  (7), 

dz  =  y  dx  -\-  X  dy. 

The  first  term  on  the  right-hand  side  represents  the  area  of  the 
strip  BEFC,  Fig.  125.     The  second  term  the  area  of  DCGH.     The 

difference  between  Az  and  dz  is  the 
^     area  of  the  rectangle   CFLG,  which 


becomes      relatively      smaller,      the 
smaller  dx  and  dy  become. 

The  above  expression  could  have 
been  obtained  by  the  formula  for  the 
B  dx  E     differential   of   the   product   of   two 
Fig.  125.  variables. 

Illustration  2.  The  base  of  a 
rectangular  piece  of  brass  is  15  feet  and  its  altitude  is  10  feet. 
If  the  base  is  increasing  in  length  at  the  rate  of  0.03  foot  per  hour 
and  the  altitude  at  the  rate  of  0.02  foot  per  hour,  at  what  rate  is 
the  area  changing? 

Let  X  denote  the  base,  y  the  altitude,  and  z  the  area. 
Then 

z  =  xy 


§171]    TOTAL  DERIVATIVE.     EXACT  DIFFERENTIAL    331 

and  dz  _     dx  dy 

dt~'^  dt  '^^  dt 

=  (10)  (0.03)  +  (15)  (0.02). 

X 


Illustration  3.     z  =  -• 

y 

^  _  1, 

dx       y 

dz 

dy 

X 

~  y'- 

and,  by  (7), 

dz  =  —  dx 5  dy, 

y  y^    ^' 

or 

,        y  dx  —  x  dy 

dz  =  5 > 

yi 

a  result  which  could  have  been  obtained  by  differentiating  the 

X 

quotient  -  by  the  usual  rule. 

Exercises 

Find  by  formula  (7)  the  total  differential  of  each  of  the  following 
functions : 


1.  z  =  x^y. 

-=^.- 

7.  2  =  x'e". 

2.    2    =  XJ/^ 

6.  2  =  z  log  y. 

S.  z  =  e*  sin  y. 

a;2 
3.  2  =  — 

y 

6.  z  =  e*  cos  y. 

9.  2  =  e""  cos  nx. 

Find  ^f  if: 
dt 

10.  z  =  X*  cos 

y- 

11.   2 

=  e* 

sin  y. 

12.  The  radius  of  the  base  of  a  right  circular  cylinder  is  8  inches  and 
its  altitude  is  25  inches.  If  the  radius  of  the  base  is  increasing  at 
the  rate  of  0.2  inch  per  hour  and  its  altitude  at  the  rate  of  0.6  inch 
per  hour,  at  what  rate  is  the  volume  increasing? 

13.  Given  the  formula  connecting  the  pressure,  volume,  and  tem- 
perature of  a  perfect  gas,  pv  =  Rt,  R  being  a  constant.  11  t  =  523°, 
p  =  1500  pounds  per  square  foot,  and  v  =  21.2  cubic  feet,  find  the 
approximate  change  in  p  when  I  changes  to  525°  and  v  to  21.4  cubic 
feet. 

14.  If  with  the  data  of  Exercise  13,  the  temperature  is  changing  at 


332  CALCULUS  [§172 

the  rate  of  1°  per  second,  while  the  volume  is  changing  at  the  rate  of 
0.4  cubic  foot  per  second,  at  what  rate  is  the  pressure  changing? 

15.  The  edges  of  a  rectangular  parallelopiped  are  6,  8,  and  10  feet. 
They  are  increasing  at  the  rate  of  0.02  foot  per  second,  0.03  foot  per 
second  and  0.04  foot  per  second,  respectively.  At  what  rate  is  the 
volume  increasing? 

172.  Exact  Differential.     An  expression  of  the  form 

M  dx  -{-  N  dy, 

where  M  and  N  are  functions  of  x  and  y,  may  or  may  not  be  the 
differential  of  some  function  of  x  and  y.  If  it  is,  it  is  called  an 
exact  differential.     Thus 

sin  y  dx  -\-  X  cos  y  dy  (1) 

is  an  exact  differential,  for  it  is  the  differential  of  z  =  x  sin  y. 

The  coefficient  of  dx  is  ^  =  sin  y,  and  that  of  dy  is  \-  =  x  cos  y. 

x^  sin  y  dx  -\-  x  cos  y  dy  (2) 

is  not  an  exact  differential.     It  is  fairly  evident  from  (1)  that  we 

dz 
cannot  find  a  function  z  =  f{x,  y)  such  that  v-  =  x^  sin  y  and 

dz 

-^  —  X  cos  y. 

dy  ^ 

We  seek  to  find  a  test  for  determining  whether  or  not  an  ex- 
pression of  the  form 

Mdx  +  N  dy  (3) 

is  an  exact  differential.  If  (3)  is  the  exact  differential  of  a  func- 
tion z,  we  must  have, 

|  =  M  (4) 

and 

1  =  ^.  (5) 

since 


§172]    TOTAL  DERIVATIVE.     EXACT  DIFFERENTIAL   333 

Differentiate  (4)  with  respect  to  y  and  (5)  with  respect  to  x 
and  obtain 

AU  AM 

(7) 
(8) 


dh         dM 

and 

dy  dx  ~~   dy 
d^z         dN 

Since,  in  general, 

dxdy  "  dx 
d^z          dh 

dydx  ~  dxdy 

it  follows  that  if  (3)  is  an  exact  differential,  we  must  have 

dy  dx 
The  condition  (9)  must  be  satisfied  if  (3)  is  an  exact  differential. 
It  does  not  follow,  however,  without  further  proof,  that  (3)  is 
an  exact  differential  if  (9)  is  satisfied.  It  can,  however,  be  shown 
that  this  is  the  case.  The  proof  will  be  omitted.  (3)  cannot  be 
an  exact  differential  unless  (9)  is  satisfied  and  is  an  exact  differ- 
ential if  (9)  is  satisfied. 

When  an  expression  of  the  form  (3)  is  given,  the  first  step  is  to 
determine  whether  or  not  it  is  an  exact  differential  by  applying 
the  test  (9).  If  it  is  an  exact  differential,  the  next  step  is  to 
find  the  function  z  of  which  it  is  the  differential.  This  step  will 
be  illustrated  by  integrating  several  differentials  for  which  the 
functions  from  which  they  were  obtained  by  differentiation  are 
known. 

Illustration  1.     If  z  =  a;^  +  2xhj  -\-  y^  -\-  C, 

^'  =  dic'^''  +  dl^y 

=  {3x^  +  4xy)dx  +  (2x^  +  2y)dy. 
If  then  we  are  given  the  exact  differential 

dz  =  (3a;2  +  ^xy)dx  +  {2x^  +  2y)dy 
and  arc  required  to  find  the  function  of  which  it  is  the  differential, 
we  note  first  that 


dz 
Then 


d^x  =  ^^'  +  ^^2/. 


2  =  a;3  _|_  2xhj  +  a  Junction  of  y  alone. 


334  CALCULUS  [§172 

And  this  function  of  ?/  is  to  be  so  determined  that 
dz 

Clearly  the  term  2x^  is  obtained  by  taking  the  derivative  with 
respect  to  t/  of  2x^2/,  a  term  already  found,  and  consequently  it 
is  not  to  be  added.  2y  is  the  derivative  of  y^.  y^  is  then  the 
function  of  y  which  is  to  be  added  to  the  terms  already  found. 
Further  an  arbitrary  constant  is  to  be  added  since  its  differential 
will  be  zero.     Then 

z  =  x'  +  2xh/  +  y^  +  C 

is  the  function  whose  differential  was  given.  If,  as  is  usually  the 
case,  it  had  been  given  that 

(3x*  +  ^xy)dx  +  (2x2  +  2y)dy  =  0  (10) 

it  would  have  been  required  to  find  a  function  of  x  and  y  such  that 
its  differential  would  be  zero.  Now  the  first  member  is,  as  we 
have  seen,  the  differential  of 

z  =  x^  +  2xh)  +  y\ 
But,  if  (fe  =  0,  2  =  C.    Then 

x»  +  2xhi  +  y^  =  C 

is  the  relation  between  x  and  y  which  satisfies  the  given  equation. 
Illustration  2.     If 

z  =  e*  cos  y  -{-  x^  -\-  sin  y  +  y^, 

dz  =  (e*  cos  y  -\-  2x)  (fx  +  ( —  e*  sin  y  +  cos  y  +  ^y^)  dy 

Now  let  it  be  given  that 

(e*  cos  y  +  2x)  dx  -\-  (—  e'  sin  y  -\-  cos  y  +  Sy^)  dy  =  0.  (11) 

From  its  derivation  we  know  that  the  left-hand  member  is  an 
exact  differential,  dz.  Let  us  proceed  to  find  z  as  if  it  were 
unknown. 


dz 
Then 


-—  =  e"  cos  y  +  2x. 


z  =  e*  cos  y  +  x^  +  a  function  of  y  alone.  (12) 


§173]    TOTAL  DERIVATIVE.     EXACT  DIFFERENTIAL    335 

The  function  of  y  is  to  be  so  determined  that 

dz 

^  =  -  e' sm  y  +  cos  rj  -{■  dy'^.  (13) 

The  first  term  is  evidently  obtained  by  differentiating  e'  cos  y,  a 
term  already  found  in  (12).     The  remaining  two  terms  in  (13) 
are  obtained  by  differentiating  sin  y  +  y^.     These  are  to  be  added 
to  the  terms  already  found  in  (12). 
Then 

z  =  e'  cos  y  -{-  x^  -\-  sin  y  +  y^. 
But,  since  dz  =  0,  z  =  C.     Hence 

e"  cos  y  -{-  x^  -{-  sin  y  -\-  y^  =  C 

is  a  solution  of  (2). 

Illustration  3,     Integrate  if  possible  the  equation 

{e'  y+sin  y-\-2x)  da;+(e'+x  cos  y-\-e''-\-2y—sin  y)  dy  =  0.   (14) 

We  have  first  to  determine  whether  or  not  the  first  member  is 
an  exact  differential.     Apply  the  test  (9), 

dM 

~  =  e'  +  cos  y. 

dN 

——  =  e*  +  cos  y. 

Hence  (9)  is  satisfied  and  the  first  member  of  (14)  is  an  exact 
differential.  On  integrating  the  coefficient  of  dx  with  respect  to 
X  we  obtain 

c^y  -{-  xsiny  -{-  x"^. 

To  this  we  have  to  add 

gV    ^   y2    _|_    pQg  y^ 

the  terms  which  arise  from  the  integration  of  the  coefficient  of  dy 
and  which  contain  y  alone.  (The  other  terms  in  the  coefficient  of 
dy  arise  from  the  differentiation  of  terms  already  found  by  integrat- 
ing the  coefficient  of  dx.)     Then  the  solution  of  (14)  is 

e^y  +  xsiny  +  x^  +  e"  +  y^  +  cosy  =  C. 

173.  Exact  Differential  Equations.  Equations  involving  differ- 
entials or  derivatives  are  called  differential  equations.  Those  of 
the  type 

Mdx  +  N  dy  =  0  (1) 


336  CALCULUS  [§174 

where  the  first  member  is  an  exact  differential,  are  called  exact 
differential  equations. 

The  equations  (10),  (11),  and  (14)  of  Illustrations  1,  2,  and  3, 
§172,  are  exact  differential  equations.  The  process  of  finding 
the  relation  between  y  and  x,  which  when  differentiated  gives  a 
certain  differential  equation,  is  called  the  integration  of  the 
equation. 

The  procedure  in  dealing  with  an  equation  of  type  (1)  is  to 
determine  first  whether  or  not  it  is  exact  by  applying  the  test  (9), 
§172.  If  it  is,  integrate  the  coefficient  of  dx  with  respect  to  x 
and  to  this  result  add  those  terms  which  contain  y  only,  which 
are  obtained  by  integrating  the  coefficient  of  dy  with  respect  to  y. 

Exercises 

Are  the  following  differential  equations  exact?  Integrate  those 
which  are  exact. 

1.  Sx'^y^dx  +2xhjdy  =  0. 

2.  —  cos(-)dx cos (-)  dy  =  0. 

y       \yl         y'      \y/ 

3.  y  e'"  (1  +  X  +  y)  dx  +  X  e'"  {I  +  x  +  y)  dy  =  0. 

4.  y  e'"  dx  +  X  e'"  dy  =  0. 

5.  {x^y  +  2x)  dx  -  {Sx^y  -  5x)  dy  =  0. 

6.  (p+l)dx-  [^  +  2y)dy=0. 

7.  e-y(2  + -^^  -  j^    e^2  +  ^^dy=0. 

174.  In  §155  the  envelope  of  a  family  of  curves  was  defined, 
and  its  parametric  equations  were  found  to  be 

fix,  y,  c)  =0  (1) 

We  shall  now  show  that  the  envelope  is  tangent  to  each  curve 
of  the  family  of  curves  (1). 

At  a  given  point  {x,  y)  of  the  curve  determined  by  giving  c 
a  particular  value  in  (1),  the  slope  of  the  tangent  is  found  from 
the  equation 

dj.dfdj,^^ 
dx       dy  dx         ' 


§174]    TOTAL  DERIVATIVE.     EXACT  DIFFERENTIAL    337 

If  the  point  also  lies  upon  the  envelope  its  coordinates  satisfy 

(1)  and  (2).     The  equation  of  the  envelope  can  be  regarded  as 
given  by  (1)  where  c  is  the  function  of  x  and  y  found  by  solving 

(2)  for  c.     On  differentiating  (1)  with  respect  to  x,  regarding  c 

as  a  function  of  x  and  y,  the  slope,  -r-,  of  the  tangent  to  the  en- 
velope is  given  by 

dx       dy  dx       dc    dx         ' 

dc  _  dc       dc   dy 
dx       dx       dy  dx 


where 


fif 

But  on  the  envelope  ^  =  0.     Hence  (4)  becomes 

^  +  ^  ^  =  0  (5) 

dx       dy  dx         ' 

Equations  (3)  and  (5)  show  that  the  slope  of  the  tangent  line  to 
the  envelope  at  the  point  (x,  y)  is  the  same  as  the  slope  of  the  tan- 
gent line  at  the  same  point  to  a  curve  of  the  family  (1).  Hence 
the  envelope  is  tangent  to  each  curve  of  the  family  of  curves  (1). 


22 


CHAPTER  XIX 
DIFFERENTIAL  EQUATIONS 

175.  Differential  Equations.  An  equation  containing  deriva- 
tives or  differentials  is  called  a  differential  equation.  If  no  deriva- 
tive higher  than  the  first  appears  it  is  called  a  differential  equation 
of  the  first  order.  If  the  equation  contains  the  second,  but  no 
higher  derivative,  the  equation  is  said  to  be  of  the  second  order. 
And  so  on.  Numerous  differential  equations  have  already 
occurred  in  this  course.  We  shall  now  consider  the  solution  of 
differential  equations  somewhat  systematically. 

176.  General  Solution.     Particular  Integral.     Let 

fix,  y,c)=0  (1) 

be  any  equation  between  x,  y,  and  the  constant  c.     If  (1)  is 
differentiated  with  respect  to  x  there  results  the  equation 

Fix,  y,  y',  c)  =  0.  (2) 

Between  (1)  and  (2)  the  constant  c  can  be  eliminated  giving  the 
differential  equation  of  the  first  order 

<l>{x,  y,  y')  =  0.  (3) 

Equation  (3)  follows  for  any  value  of  the  constant  c. 

Let 

/(x,  y,  ci,  C2)  =  0  (4) 

be  an  equation  involving  two  constants,  Ci  and  Co.     By  differ- 
entiating (4)  we  obtain 

Fix,  y,  y',  ci,  d)  =  0  (5) 

and 

<t>ix,  y,  y',  y",  ci,  c^)  =  0.  (6) 

Between  equations  (4),  (5),  and  (6),  Ci  and  C2  may  be  eliminated 
giving  the  differential  equation  of  the  second  order 

Hx,  y,  y',  y")  =  0.  (7) 

338 


§178]  DIFFERENTIAL  EQUATIONS  339 

From  the  equation  (1)  containing  one  arbitrary  constant  the 
differential  equation  of  the  first  order  (3)  is  obtained.  From  the 
equation  (4)  containing  two  arbitrary  constants  the  differential 
equation  of  the  second  order  (7)  is  obtained.  In  like  manner 
from  a  relation  between  x  and  y  containing  n  arbitrary  constants 
a  differential  equation  of  the  n*^  order  is  obtained  by  differentiat- 
ing, and  eliminating  the  constants. 

Equation  (1)  is  a  solution  of  equation  (3).  It  is  called  the 
general  solution  and  involves  one  arbitrary  constant  of  integration, 
c.  Equation  (4)  is  called  the  general  solution  of  (7).  It  involves 
two  arbitrary  constants,  or  constants  of  integration.  It  can  be 
shown  that  the  general  solution,  or  general  integral,  of  a  differential 
equation  contains  a  number  of  arbitrary  constants,  or  constants 
of  integration,  equal  to  the  order  of  the  differential  equation. 

A  particular  integral  is  obtained  from  the  general  integral  by 
giving  particular  values  to  the  constants  of  integration. 

177.  Exact  Differential  Equations.  This  type  of  differential 
equation  was  discussed  in  §173. 

178.  Differential  Equations;  Variables  Separable.  The  vari- 
ables X  and  y  are  said  to  be  separable  in  a  differential  equation 
which  can  be  put  in  the  form  f(x)  dx  +  </>(?/)  dy  =  0.  The  first 
member  is  equal  to  a  function  of  x  alone  multiplied  by  dx  plus  a 
function  of  y  alone  multiplied  by  dy. 

Illustration  1. 

(1  +  y'^)x  dx-\-  {l-\-  x^)y  dy  =  0. 

On  dividing  by  (1  +  ?/^)(l  +  x^)  this  equation  becomes 

xdx  y  dy    _ 


l  +  x^   '    14-2/ 
Integration  gives 

ilog  (1  +  x^)  +  ilog  (1  +  2/2)  =  C. 

This  reduces  to 

'1> 


(1  +  x^){\  +  2/2)  =  e2C  =  c. 


or 


340  CALCULUS  I§179 

Illustration  2. 

Vl  -  y^  dx  +  \/l  -  x2  rf?/  =  0. 
Then 

dx  dy        ^  ^ 

Vl  -ic2       Vl  -  2/'         ' 

and  the  variables  are  separated.     Integration  gives 

sin~i  X  +  sin~i  y  =  C. 

Take  the  sine  of  each  member,  observing  that  the  first  member  is 
the  sum  of  two  angles,  and  obtain 

a;Vl  -  2/'  +  yVi  -  x""  =  sin  C  =  d. 

Exercises 

Solve  the  following  differential  equations: 

1.  (1  -  x)dy  -  (1  +  y)dx  =  0.  Ans.  (1  +  y){l  -  x)  =  C. 

2.  sin  X  cos  y  dx  =  cos  x  sin  y  dy. 

8.  (x  -  Vl  +  x^)  Vl  +  2/^  dx  =  (1  +  x^)dy. 

4.  ^^  =  5y^x. 

f.    dy       y"  +  4?/  4-  5  _ 
**•  dx  "^  x2  +  4x  +  5  ~ 

6.  (1  +  a:)dj/  =  y(l  —  i/)dx.         ^ns.     ?/  =  c(l  +  x)(l  —  y). 

7.  (1  -  x)ydx  +  (1  +  j/)x  dy  =  0. 

dv 

8.  ^  +  e'2/  =  e'y«. 

9.  (x»  +  j/x*)di/  -  (t/2  -  x?/2)dx  =  0. 

dy   ,   ^  dy 

11.  3e*  sin  y  dx  +  (1  —  e')  cos  y  dy  =  0. 

12.  (xy  +  x»2/)dy  -  (1  +  y^)dx  =  0. 

Ans.     (1  4-  x')(l  +  y')  =  ex*. 

179.  Homogeneous   Differential   Equations.     The   differential 
equation 

Mdx-\-N  dy  =  Q  (1) 

is  said  to  be  homogeneous  if  M  and  iV  are  homogeneous  functions 
of  X  and  y  of  the  same  degree. 


§179]  DIFFERENTIAL  EQUATIONS  341 

A  function  f{x,  y)  of  the  variables  x  and  y  is  said  to  be  homogen- 
eous of  degree  n  if  after  the  substitutions  x  =  Xx',  y  =  \y'  have 
been  made, 

fix,  y)  =  X-/(x',  y'). 
Thus 

ax^  +  bxy  -\-  cy^ 

is  liomogeneous  of  degree  2.     For,  on  making  the  substitutions 
indicated,  it  becomes 

\\ax'^  +  bx'y'  +  cy'^). 
The  expression 


ax^Vx^  +  1/2  +  6x3  tan-»  (-\ 


is  homogeneous  of  degree  3.  For,  after  the  substitutions  indi- 
cated above,  it  becomes 

X3  fox'^^/x'^  +  y'^  +  6x'3  tan-i  Z^;)  ]  • 

A  homogeneous  differential  equation  of  the  form  (1)  is  solved  by 
placing  y  =  vx,  and  thus  obtaining  a  new  differential  equation 
in  which  the  variables,  v  and  x,  are  separable. 

Illustration: 

(x2  +  2/2)  dx  +  3xy  dy  =  0. 
Let 

y  =  vx. 
Then 

dy  =  V  dx  -\r  X  dv, 
and 

x2(l  +  v^)  dx  +  Svx%v  dx  -\-x  dv)  =  0. 
x2(I  +  4«;2)  dx  +  3^3  dv  =  0. 


Separating  the  variables 


dx         Svdv     _ 
T  "•"  1  +  4i;2  "  "• 
log[x(l  +4i;2)i]  =  C, 

x(l  4-4i'2)i  =  Ci. 


342  CALCULUS  [§180 

y 

On  substituting  t;  =  -  we  obtain  as  the  solution  of  the  given 
equation 

or 

x\x^  +  4?/2)3  =  Ci. 

Exercises 

Solve  the  following  differential  equations 

^  _  ^„  #. 
dx  ~  ^y  dx 

2.  x'^y  dx  -  (x'  +  2/3)  di/  =  0. 

3.  (8y  +  10a;)  dx  +  (5y  +  7x)  dy  =  0. 

4.  (2\/x?/  —  x)  dy  +  2/  dx  =  0. 

?/  ^2/  2/ 

6.  X  cos  —  -J-  =  y  cos X. 

X  dx  X 

dy 


7.  x  ^  -  y  =  Vx''  -  y\ 

8.  {y  -  x)  dy  +  ydx  =  0. 

180.     Linear  Differential  Equations  of  the  First  Order.     The 

equation 

where  P  and  Q  are  functions  of  x  only,  is  called  a  linear  differential 
equation.  It  is  of  the  first  degree  in  y  and  its  derivative.  Multi- 
ply the  equation  by 

'  Pdx 


and  obtain 


./' 


iPdxrdy    ,    „    1  {pdx   ^ 


(2) 


The  left-hand  member  is  the  derivative  of 

fpdx 
eJ         y, 

as  may  be  confirmed  by  differentiating  this  product.     The  inte- 
gration of  (2)  gives 


}180] 


DIFFERENTIAL  EQUATIONS  343 


e* 

f'-"v  =  JQeJ'""dx  +  C. 

Illustration  1. 

t+^y-"'- 

Here  P  =  x  and  Q 

=  x\     Then 

I  Pdx             \  xdx          T 

X- 

Multiply  both  members  of  (3)  by  c  2 . 

Integration  gives 

a;2          /.    a;2 

a;2                 a;2 

=  e^x^  -  2e2  +  C. 

Hence 

a;2 

J/  =x2-2  +  Ce~"^. 

Illustration  2. 

I  Pdx  (^ 

oJ  =  P.  ''  = 


Multiply  both  members  of  (4)  by  x. 

Integration  gives 

xy 
or 


=  ^  +  x^*  +  2x2  +  C, 


(3) 


^  +  l^=:.2  +  3^+4.  (4) 

dx      x*^ 


x'  C* 

y  =  ^  +  x24-2x  +  -- 

This  illustration  is  inserted  to  call  attention  to  the  well-known 
simple  relation  e'"«^  =  x,  which  there  will  be  frequent  occasion 


344  CALCULUS  [§181 

CO  use  in  solving  equations  of  this  type.     It  should  be  recalled 
that  e''^°«^  =  e'°«(^">  =  x".     Thus 


e 

—  loga;  _  t 
X 

Exercises 

1. 

dy 
dx 

+  2xy  =  e 

■^\ 

2. 

dy 
dx 

+  y  cos  X  = 

sin 

2x. 

.   dy   , 

3.  cos'^x-T — r  y  =  tan  x. 

4.  (x«  +  l)^  +  2xy  =  4x2. 

'•g  +  ^  =  (-  +  «•• 

6.  x(l  -  x«)  dy  +  (2x2  _  i)^  ^^  =  ^x^  ^^.^ 

„    dy         y 
1.  -/  -n-  =  e'x". 
dx         X 

8.  (1  +  x^)  dy  ^-  (xy  -^  dx  =  Q. 

^    dy   ,1  -2x 

10.  (1  +  y"^)  dx  =  (tan-iy  -  x)  dy. 

181.  Extended  Form  of  the  Linear  Differential  Equation.     An 
equation  of  the  form 

%  +  Pv  =  Qr  (1) 

is  easily  reduced  to  the  linear  form.  For,  on  dividing  (1)  by  y, 
we  obtain 

The  first  term  of  the  left-hand  member  of  (2)  is,  apart  from  a 
constant  factor,  the  derivative  of  ?/""+',  which  occurs  in  the  second 
term.  If  we  let  z  =  2/~"^^  we  obtain  the  linear  differential 
equation 

or 

g  _  (n  -  l)Pz  =  -in-  1)Q. 


§1821  DIFFERENTIAL  EQUATIONS  345 

Illustration  1. 


dv 

T — h  2/  cos  X  =  y*  sin  2x. 


Dividing  by  y* 
Let  2  =  7/-3.     Then 


.dy,,  .    „ 

y~*  -j~  +  y~^  cos  X  =  sin  2x, 


,-4  ^  =   _  1  ^ 


and  the  equation  becomes 
dz 
dx 


^  dz    ,  .    „ 

—  5  ;i — \-  z  cos  X  =  sin  2x, 


or 

3 Zz  COS  X  =  —  3  sin  2a;. 

dx 

This  equation  can  be  readily  solved  by  §180;  y~^  is  to  be  sub- 
stituted for  z  in  the  result. 

Exercises 

1-  ^-  +  ~  2/  =  a;V.  4.  (1  -  x2)  ^  -  xy  =  axy^. 


dy 
2.  ;^^  +  2/  =  xy3. 

^'t+ly  =  '^'y^- 

3.3,^1-7,3  = 

X  +  1.                 6.  X  ^  +  ,  =  ,2  log  X. 

7    '^^  +      2      ,  -  ^'. 

182.  Applications.  Let  there  be  an  electric  circuit,  whose 
resistance  is  R,  whose  coefficient  of  self-induction  is  L,  and  which 
contains  an  electromotive  force,  which  at  first  we  shall  suppose 
constant  and  equal  to  E.  It  is  required  to  find  the  current  i  at 
any  time  t  after  the  time  <  =  0,  at  which  the  circuit  was  closed. 
The  equation  connecting  the  quantities  involved  is  readily  set  up. 
The  applied  E.M.F.,  E,  must  overcome  the  resistance  of  the  circuit 
and  its  self-induction.  The  former  requires  an  E.M.F.  equal  to 
iR,  and   the  latter  an  E.M.F.  proportional  to  the  time  rate  of 

dz  dz 

change  of  current,  viz.,  j,  and  equal  to  L  ^.     The  applied  E.M.F., 

E,  must  equal  the  sum  of  these  two  E.M.F.'s. 


346  CALCULUS  [§182 

Hence  ^  di   ,    ^.       ^  ,,. 

Lj^  +  Ri  =  E,  (1) 

The  student  will  show  that,  if  i  equals  zero  when  t  equals  zero, 
the  solution  of  this  linear  equation  is 

Er  ^h 

If  the  battery  or  other  source  of  E.M.F.  is  suddenly  cut  out  of 
the  circuit,  the  current  falls  off  in  such  a  way  that  the  differential 
equation 

Lj^  +  Ri  =  0  (3) 

is  satisfied.     Show  that  the  law  at  which  the  current  falls  off  is 

t  =  ioe-L^'-  '»>,  (4) 

if  the  instant  at  which  the  battery  is  cut  out  is  the  time  1  =  1^ 
and  if  the  current  at  this  instant  is  i  =  Iq. 

If  the  E.M.F.  is  variable,  the  relation  between  the  quantities 
involved  in  the  circuit  is  still  governed  by  (1), 

Lj^  +  Ri^E,  (1) 

in  which  E  is  now  variable.  Suppose  E  =  Eo  sin  cof.  This  sup- 
poses that  an  alternating  E.M.F.  is  acting  in  the  circuit.  The 
differential  equation  to  be  solved  is 

di 
L-r.  +  Ri  =  Eo  sin  cat.  (5) 

Show  that 

*«       ~  L  R     , 


R^  +  co^L 


y-  sin  cat  —  03  cos  coHe  ^     +  C 

ft 
(R  sin  cat  —  caL  cos  cat)e       +  C 


Eq  .  ,        .  , X        ^ 


^f 


V<R'  +  w^L^ 


=r-rr-  siu  (cof    —    0)e  +   C, 


§183]  DIFFERENTIAL  EQUATIONS  347 

where 

sm  <f)  = 


Then, 


cos  d>  =  — ,-  =. 


R 


Jp  — ^  t 

i  =      ^       °  sin  (co<  -  0)  +  Ce  ^ 


Since  the  last  term  becomes  negligible  after  a  short  time  because 
of  the  factor 

e  ^    > 

it  is  scarcely  necessary  to  determine  C.  On  droppmg  out  the 
last  term  as  unimportant  except  in  the  immediate  vicinity  oit  =0, 
we  have 

The  current,  therefore,  alternates  with  the  same  frequency  as  the 
E.M.F.,  but  lags  behind  it  and  differs  from  it  in  phase  by  4>-     It 

is  to  be  noted  that  the  maximum  value  of  the  current  is  not  -^5- 

K 
JP 

but      .  •     The  quantity  -s/R^  +  oj^L^  replaces,  in  alter- 

V /2^  +  co^L^ 

nating  currents,  the  resistance  R  of  the  ordinary  circuit.     It  is 

called  the  impedance  of  the  circuit. 

183.  Linear  Differential  Equations  of  Higher  Order  with  Con- 
stant CoeflScients  and  Second  Member  Zero.  A  typical  differen- 
tial equation  of  this  class  is  the  following: 

d"y   ,        d^-^y   ,        d"~hj   ,  ,  dy   ,  ^    /,s 

where  Oq,  Oi,  •  •  •,  a„  are  constants.  As  the  equations  of  this 
class  which  occur  in  the  applications  are  usually  of  the  second  order 
we  shall  confine  our  discussion  in  this  article  to  linear  differential 
equations  of  the  second  order.     Consider 


348  CALCULUS  [§183 

Let  us  assume  that 

2/  =  e-"*  (3) 

and  find,  if  possible,  the  values  of  m  for  which  (3)  is  a  solution  of 
(2).     The  substitution  of  (3)  in  (2)  gives 

e""  {aom^  +  aim  +  a^  =  0.  (4) 

The  first  factor  cannot  vanish.  The  second,  equated  to  zero,  gives 
a  quadratic  equation  in  m.  Call  its  roots  nii  and  mt.  Then  (3) 
is  a  solution  of  (2)  if  m  has  either  of  the  values  Wi  or  mo,  the  roots 
of 

OqW^  +  aim  +  a2  =  0.  (5) 

The  equation  (5)  in  m,  obtained  from  the  given  differential  equa- 

d'^v  du 

tion  by  writing  m^  for  y-^  and  to  for  -v-  is  called  the  auxiliary 

equation. 
Two  solutions  of  (2)  are 

y  =  gmii     and     y  =  e'^i*. 
Furthermore, 

y  =  Cie-"!" 

is  a  solution  of  (2).  For,  after  the  substitution  of  this  value  of 
y  in  (2),  Ci  can  be  taken  out  as  a  common  factor  and  the  other 
factor  vanishes  in  accordance  with  (4)  or  (5).     In  the  same  way, 

y  =  de'^i' 

is  a  solution  of  (2).     And  finally  the  sum  of  the  two  solutions 

y  —  Cie""!^  +  C2e"'2*  (6) 

is  a  solution  of  (2).  This  can  be  seen  by  substituting  in  (2)  and 
recalling  that  nii  and  mo  are  roots  of  (5).  When  mi  is  not  equal 
to  m2,  (6)  is  known  as  the  general  solution  of  the  differential  equa- 
tion (2).  It  contains  two  arbitrary  constants,  the  number  which 
the  general  solution  of  a  differential  equation  of  the  second  order 
must  contain. 

The  values  of  these  constants  are  determined  in  a  particular 
problem  by  two  suitable  conditions. 

Illustration. 


§184]  DIFFERENTIAL  EQUATIONS  349 

The  auxiliary  equation  is 

m^  —  bm  +  6  =  0, 
[m  -  2){m  -  3)  =  0. 

Hence  mi  =  2,  mz  =  3.     The  general  solution  is  then 

y   =   Cie^'  +  C26''. 


Exercises 

»-g-l-''-°- 

2.g-4,  =  0. 

»-g-4^'»-»- 

*-g  +  |->^-»- 

'■2+^t-o- 

184.  Auxiliary  Equation  with  Equal  Roots.  The  method  just 
given  fails  when  the  auxiliary  equation  has  equal  roots,  mi  =  W2. 
For,  equation  (6),  §183,  becomes 

y  =  de""!'  +  C2e'"2* 
=  (Ci  +  C2)e-^'. 

But  Ci  +  C2  is  an  arbitrary  constant  and  the  solution  contains 
only  one  arbitrary  constant  instead  of  two.  When  the  auxiliary 
equation  has  equal  roots,  mi  =  7^2,  equation  (2)  can  be  written 
in  the  form 

Its  general  solution  is 

This  solution  can  be  verified  by  direct  substitution. 
Illustration. 

^_4^+4v  =  0. 
dx^         dx^    y      "• 

The  auxiliary  equation  is 

m^  —  4m  +  4  =  0. 
TMi  =  m2  =  2. 


350  CALCULUS  [§185 

The  general  solution  is 

2/  =  (C,  +  C^x)e^'. 

Exercises 

l-^^  +  '^g  +  o-"- 

185.  Auxiliary  Equation  with  Complex  Roots.  If  the  auxiliary 
equation  has  complex  roots  the  general  solution  can  be  written 
in  a  form  different  from  (6),  §183.  The  importance  of  the  result 
will  be  evident  at  once  when  it  is  observed  that  it  contains  the 
harmonic  functions  sine  and  cosine.  If  the  coefficients  of  the 
given  differential  equation  (2),  §183  are  real,  and  if  mi  and  niz 
are  complex,  they  must  be  conjugate  imaginary  numbers.  Let 
THi  =  a  -\-  ih.     Then  mz  =  a  —  ih.     Then  (6)  becomes 

y  =  Cie"  +  '*'  +  Cze  "'  ~  *'" 

Now,  by  (4)  and  (5),  §167, 

gift*  =  cos  bx  +  i  sin  bx 

Q-ibx  —  COS  bx  —  i  sin  bx. 
Then 

y  =  c*  [(Ci  +  C-i)  cos  bx  +  i{Ci  —  Cz)  sin  6a;] 

On  placing  Ci  +  d  =  A  and  i{Ci  —  Cz)  =  B,  we  obtain 
y  =  go*  (^  cos  bx  -\-  B  sin  bx) 
=  e"'  C  cos  (6a;  —  <f>). 

In  the  last  form  the  two  arbitrary  constants  of  integration  are 
C  and  <f). 
Illustration  1. 

The  auxiliary  equation  is 

m2  +  4m  +  13  =  0. 


§186]  DIFFERENTIAL  EQUATIONS  351 

I       Hence 

m  =  -2  ±  Si. 
Then 

y  =  e-2x  (J^  COS  Sx  -\-  B  sin  3a;). 

Illustration  2. 

2^  +  ^y  =  o. 


Whence 
Then 


m^  +  4  =  0. 
m  =  ±  2i  =  0  +  2i. 
T/  =  A  cos  2x  +  i5  sin  2x. 
Exercises 


d^ 
d^9 

5^  +  .^.  =  o. 


2.^,  +  9y  =  0. 


dx^+'^  =  0- 


186.  Damped  Harmonic  Motion.  The  resistance  offered  by  the 
air  to  the  motion  of  a  body  through  it,  is  roughly  proportional  to 
the  velocity,  if  the  velocity  is  a  moderate  one.  In  §81,  the  differ- 
ential equation  of  the  motion  of  the  simple  pendulum  was  derived 
on  the  assumption  that  the  force  of  gravity  was  the  only  force 
acting  upon  the  bob  of  the  pendulum.  If  the  resistance  of  the  air 
is  also  taken  into  account  we  shall  have  to  add  to  the  second 

member  of  the  equation,  I  -xr^  =  —  ^  sin  0,  a  term,  —  2kl  -r.  pro- 

portional  to  the  velocity  I  -rr-    See  equation  (1),  §81.     The  differ- 
ential equation  of  the  motion  is  then 

idW  .    ^       ^,^dd  ... 

l^=-g.me-2kl-^.  (1) 

The  negative  sign  is  used  before  the  last  term  because  the  force 
due  to  the  resistance  of  the  air  acts  in  a  direction  opposite  to  that 


352  CALCULUS  I  §186 

of  the  motion.     The  advantage  of  choosing  2k  as  the  proportion- 
ality factor  instead  of  k  will  appear  later.     A;  is  a  positive  constant. 
From  (1)  we  obtain 

j;|+2.f  +  f.„.  =  0.  (2) 

As  in  §81  assume  that  6  is  small  and  replace  sin  6  by  9.    Also  let 

g 

I 

d^e  dd 

This  is  a  linear  differential  equation  of  the  second  order  with 
constant  coefficients  and  can  be  solved  by  the  method  of  §185. 
The  auxiliary  equation  is 

m^  +  2km  +  w^  =  0, 
whence 

m  =  -  k  ±  \/k'^  -  o)  2. 

When  the  velocity  is  not  very  great,  as  in  the  case  of  an  ordinary 
pendulum,  k  is  very  small  for  air  and  is  much  less  than  oj.  The 
expression  under  the  radical  sign  is  negative.     We  write  then 

m  =  -  k  ±  i  Vw^  -  A;2, 
oj^  —  k^  being  positive. 
The  solution  of  (3)  is 

e  =  Ae-*'  cos  [t^/(a^  —  k'^  -  e], 
or,  multiplying  both  sides  by  I  and  replacing  Al  by  B, 
s  =  5e~*'  cos[t\/(t)^  —  k-  —  c]. 

The  motion  is  a  damped  harmonic  motion.     The  amplitude  de- 

2t 

creases  with  the  time.     The  period      . —  is  a  little  greater 

■yct}^  —  k'^ 

2x  . 

than  — ,  the  period  of  the  free  motion. 

Since  k  is  very  small  in  comparison  with  w,  we  can,  for  an 
approximate  solution  of  our  problem,  neglect  k^  in  comparison 
with  co^.     Equation  (4)  becomes 

s  =  5e-*'  cos  (aj<  —  e).  (5) 

This  represents  the  motion  with  a  high  degree  of  approximation. 


§186]  DIFFERENTIAL  EQUATIONS  353 

The  arbitrary  constants  B  and  e  can  be  determined  by  suitable 
initial  conditions.     For  example,  let  it  be  given  that  s  =  sq  and 

ds 

'77  =  0  when  t  =  0.     On  differentiating  (5)  we  obtain 

ds 

jT=  Be-'"[  —  k  cos  (cjt  —  e)  —  CO  sin(co<  —  e)].  (6) 

For  f  =  0  we  obtain  from  (5)  and  (6) 

So  =  B  cos  € 
0  =  B(  —  k  cos  e  +  w  sin  e). 

From  the  latter  of  these  two  equations 

tan  €  =  — . 

CO 

From  the  former 


1  -\ — i  = 


to  the  degree  of  approximation  used  above.     We  have  then  as  the 
approximate  equation  of  motion 

s  =  SqC"*'  cos  (_ut  —  e)  (7) 

where 

k       k 
€  =  tan~'  —  =  — J  approximately. 

Since  k  is  very  small,  c  is  very  small. 

It  follows  from  (5)  and  (7)  that  the  period  of  the  pendulum 
in  the  case  just  considered  is  very  little  different  from  that  of 
the  same  pendulum  swinging  in  a  vacuum.  The  amplitude  of 
the  swing,  however,  is  affected  and  diminishes  continually  with 
the  time. 


23 


INDEX 

(Numbers  refer  to  pages) 


Acceleration,  52,  83, 136,  138 

angular,  138 

average,  52 
Algebraic  function,  4,  55 

definition  of,  56 

rational,  6 
Alternating  series,  316 
Angle, 

between  two  lines,  321 

between  two  planes,  236 
Angular  acceleration,  138 

velocity,  138 
Anti-derivative,  46,  49 
Applications,  218 
Arc, 

differential  of  length  of,  100, 
181 

length  of,  218 
Area, 

by  double  integration,  253, 
258 

moment  of  inertia  of,  281 

polar  coordinates,  183,  218 

of  surface  of  revolution  of, 
112,  219,  274 

under  a  curve,  75,  218 
Arithmetic  mean,  116 
Axes,  coordinate,  228 

Base, 

change  of,  147 
naperian,  146 
natural,  146 


147 

199 


func- 


Cable,  parabolic,  82 
Catenary,  161 


Center, 

of  curvature,  290,  294 

of  gravity,  250,  254,  274 

of  mass,  263 
Centroid,  263,  265,  274,  281 

of  a  line,  268 

of  an  area,  268 
•    of  a  solid,  268 
Change  of  base,  logarithm, 

of  limits  of  integration, 
Circle, 

curvature  of,  290 

of  curvature,  290 
Circular   and   exponential 

tions,  relation  between. 
321 
Circular  functions,  122,  321 
Comparison  test,  316 
Complex  numbers,  323 
Compound  interest  law,  155 
Computation 

of  IT,  321 

of  logarithms,  320 
Concavity  of  curve,  90 
Conicoid,  242 
Constant,  1 

Continuous  function,  31 
Contraction  of  curve,  7 
Convergence  of  series,  315 
Convergent  series,  303 
Co6rdinate  axes,  228 

planes,  228 
Cosines,  direction,  230 
Curvature,  289 

approximate  formula  for,  293 

center  of,  290,  294 


356 


356 


INDEX 


Curvature,  circle  of,  290 

of  a  circle,  290 

defined,  289 

parametric  equations,  292 

radius  of,  290 
Curve, 

contaction  of,  7 

direction  of,  178 

elongation  of,  7 

orthographic    projection    of, 
7 

shear  of,  7 

translation  of,  7 
Curves 

of  hyperbolic  type,  2 

maxima  and  minima  points 
of,  20 

of  parabolic  type,  2 
Cylindrical  surfaces,  244 

Damped  harmonic  motion,  351 
Definite  double  integrals,  255 

integrals,  88,  103,  104 
DeMoivre's  Theorem,  323 
Dependent  variable,  1 
Derivative,  21 

first,  21 

of  a  constant,  43 

of  a  function  of  a  function,  66 

of  a  quotient,  58 

of    circular    functions,    122, 
124,  131 

of  exponential  functions,  145 

of  logarithmic  functions,  145 

of  sin  u,  122 

of  the  product  of  two  func- 
tions, 57 

of  the  sum  of  a  function  and 
a  constant,  39 

of  the  sum  of  a  constant  and 
a  variable,  56 

of  the  sum  of  two  functions,  43 


Derivative  of  m„,  42,  59 

second,  70 

total,  328,  329 
Derivatives  of  higher  order,  par- 
tial, 247 

partial,  246 
Differential,  88,  95 

exact,  328,  332 

of    length     of     arc:     polar 
coordinates,  181,  218 
rectangular      coordinates, 
100,  218 
Differential  equation,  338 

exact,  335 

linear,  342 

of  higher  order,  347 

order  of,  338 

variables  separable,  339 
Diiferentiation,  31,  97 

implicit,  44 

logarithmic,  153 
Direction  cosines,  230,  239 

of  curve,  178 
Distance 

between  two  points,  230 

of  a  point  from  a  plane,  237 
Divergent  series,  303 
Double  integration,  251,  253,  258, 

255 
Duhamel's  theorem,  105 

Element  of  integration,  113 

Ellipsoid,  242 

Elliptic  paraboloid,  244 

Elongation  of  curve,  7 

Envelope,  289,  296,  336 
of  normals,  300 

Equation, 

differential,  338 
exact  differential,  335 
homogeneous       differential, 
340 


INDEX 


357 


Equation,  linear  differential,  342, 
347 
of  first  degree  in  x,  y  and  z, 

233 
of  a  plane,  233 

intercept  form  of,  234 
normal  form  of,  232 
Equations  of  a  line,  238 

parametric,  67 
Evolute,  289,  294,  300 

the  envelope  of  the  normals, 
300 
Exact  differential,  328,  332 

equation,  335 
Exponential  functions,  9,  145,  321 
Extended  law  of  the  mean,  307 

Falling  body,  22 
Family  of  curves,  297 
Formulas,  integration,  185,  201, 

203,  205,  206,  207 
Formulas,  Wallis',  204 
Fractions,  partial,  211 
Function,  1 

a",  9 

a  cos  X  +  '^  sin  x,  9,  165 

a  cos  X,  9 

ax'  +px  +  y,  165 

6  sin  X,  9 

sin  X,  9 

maximum  and  minimum  val- 
ues of,  20 

mx  ±  y/a"  -  x%  168 

xS  16 
Functions, 

algebraic,  4,  5,  56 

circular,  122,  321 

continuous  and   discontinu- 
ous, 31 

exponential,  145,  321 

hyperbolic,  159 

implicit,  44 


Functions,  logarithmic,  145 
power,  1 
rational,  6 

integral,  6 
transcendental,  6 
transformations  of,  10 

General  solution  of  a  differential 
equation,  338 

Harmonic  motion,  140 

damped,  351 
Homogeneous   differential  equa- 
tion, 340 
Hyperbolic  functions,  159 

paraboloid,  244 
Hyperboloid, 

of  one  sheet,  244 
of  two  sheets,  244 

Implicit  differentiation,  44 
Improper  integrals,  218,  223,  226 
Increments,  13 
Indefinite  integrals,  104 
Independent  variable,  1 
Indeterminate  forms,  30,  31,  303, 

304 
Infinite  limits  of  integration,  226 
Infinite  series,  303 
Infinitesimals,  29,  88 

limits  of  ratio  of  two,  93 

order  of,  91 
Infinity,  29 

Inflection,  point  of,  70 
Integral,  the 

I  sec'  X  dx,  204 
I  e°*  sin  nx  dx,  202 
I  e  "'  cos  nx  dx,  202 

Integrals,  improper,  218 
Integration,  46,  49,  185 


358 


INDEX 


Integration,  by  parts,  201 
double,  251,  253,  258 
formulas,     185,     201,     203, 

205,  206,  207 
of  expressions  containing, 

g-T"  +  bx  +  c,.190 

Va^  +  x^     Va^  -  x2, 
Vx^  -  a^,  196 
powers  of  x  and  of  a  +  hx, 

192 


.  Ca  smx  -\-  I 
of  I  — : 

J  c  sin  a;  +  .i 


bcos  x 


dx,  209 

J  cos  X 

of  powers   of  trigonometric 

functions,  193 
successive,  250 
triple,  250,  260 
Intercept  form  of  the  equation  of 

a  plane,  234 
Inverse  functions,  67 
Involutes,  302 

Law  of  the  mean,  306 

extended,  307 
Length  of  arc, 

polar  coordinates,  181,  218 

rectangular  coordinates,  100, 
111,  218 
Limit, 

definition  of,  27 

of     the     quotient     of     two 
infinitesimals,  93 

of  S/(x)x,  102 
Limits, 

infinite,  of  integration,  226 

of  integration,  change  of,  199 

theorems  on,  30 
Line, 

direction  cosines  of,  230 

equations  of,  238 
Linear  differential  equation, 

of  first  order,  342 

of  higher  order,  347 


Loci,  theorems  on,  10 
Logarithmic  differentiation,  153 
Logarithmic  functions,  145 
Logarithms,  computation  of,  320 

Maclaurin's  series,  311 
theorem,  303,  310 

Maxima  and  minima,  20,  60,  165 
applications  of,  174 
by  limits  of  curve,  169 
determined    by    derivative, 

169 
second    derivative  test    for, 
172 

Maximum  and  minimum  values 
of  functions,  60 

Maximum  defined,  20 

Mean,  arithmetic,  116 

Mean  value  of  a  function,  117 

Mean,  law  of  the,  306,  307 

Minimum  defined,  20 

Moment,  263 

Moment  arm,  263 

Moment  of  inertia,  250,  277 
of  area,  280,  281,  283 
of  a  solid,  284 
polar  coordinates,  283 
translation  of  axes,  280 
with  respect  to  a  plane,  285 

Naperian  base,  146 

Natural  base,  146 

Normal  form  of  the  equation  of  a 

plane,  232 
Normal,  length  of,  68 
Normals,  envelope  of,  300 

Octant,  229 

Order    of    differential   equation, 

338 
Orthographic  projection  of  curve, 

7 


INDEX 


359 


Pappus,  theorems  of,  274 
Parabolic  cable,  82 
Paraboloid, 

elliptic,  244 

hyperbolic,  244 

of  revolution,  24 
Parallel  planes,  236 
Parameter,  297 
Parametric  equations,  67,  292 
Partial  derivatives,  246 

of  higher  order,  247 
Partial  fractions,  211 
Particular  integral  of  a  differen- 
tial equation,  338,  339 
Path  of  a  projectile,  85 
Pendulum,  the  simple,  141 
Per  cent,  rate,  159 
Perpendicular  planes,  236 
Plane, 

general  equation  of,  234 

intercept  form   of  equation 
of,  234 

normal  form  of  equation  of, 
232 
Planes, 

angle  between  two,  236 

coordinate,  228 

parallel,  236 

perpendicular,  236 
Point  of  inflection,  70 
Points,  distance  between,  230 
Polar  coordinates,  178 

area,  183,  258 

centroid,  276 

differential  of  arc,  181 

direction  of  curve  in,  178 

moment  of  inertia  in,  283 
Polynomials,  4 
Power  function,  1,  33 

derivative  of,  33,  35,  41,  42 

hyperbolic  type,  2 

law  of,  3 


Power  function,  parabolic  type,  2 
Power  series,  .'505 
Projectile,  path  of,  85 

Quadric  surface,  242 
Quotient,  derivative  of,  58 

Radius  of  curvature,  290 

approximate  formula  for,  293 
Radius  of  gyration,  278 
Rate  of  change,  37 
Rational  algebraic  function,  6 
Relative  rate,  159 
Rolle's  theorem,  305 

Second  derivative,  70,  172 
Series,  303 

alternating,  316 

convergence  of,  315 

convergent,  316,  303 

divergent,  303,  315 

infinite,  303 

Maclaurin's,  311 

power,  305 

Taylor's,  311 

test  for  convergence,  315 
Shear  of  curve,  8 
Simple  harmonic  motion,  140 

pendulum,  141 
Slope  of  tangent  line,  17 
Solid  geometry,  228 
Solid  of  revolution,  109,  218 

surface  of,  219 

volume  of,  109,  218,  260 
Solution  of  differential  equation, 

338 
Subnormal,  length  of,  68 
Subtangent,  length  of,  68 
Successive  integration,  250 
Surface  of  revolution,  112,  219 
Surfaces, 

cylindrical,  244 


360 


INDEX 


Surfaces      of     revolution,     240, 
274 
quadric,  242 
Symmetric  form  of  the  equations 
of  a  line,  238 

Tangent, 

length  of,  68 

slope  of,  17 
Taylor's  series,  311 

theorem,  303,  308 
Tests  for  convergence,  316 
Theorems  of  Pappus,  274 
Total  derivative,  328 

differential,  329 
Transcendental  functions,  6 


Transformation  of  functions,  10 
Translation  of  curves,  6 
Triple  integration,  250,  260 

Variable,  1 

Velocity,  16,  52,  136,  138 

average,  16 

of  a  falling  body,  22 
Volume  by  triple  integration,  260 

of  a  solid  of  revolution,  109, 
218,  274 

Wallis'  formula,  204 
Water  pressure,  114,  219 
Work  done  by  a  variable  force, 
77,  107,  219 


Date  Due 

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inAM  oh: 

9  77 

RBTD    „, 

\h  77 

G|U 

(S 

PRINTED 

IN  U.  S.  A. 

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A     000  584  805     6 


